Proof First of all, assume item (d) and fix a decreasing sequence of dense open sets $(O_{n})_{n\in {\mathbb N}}$ . Define the nowhere dense closed set $X_{n}:= [0,1]\setminus O_{n}$ and consider $h:[0,1]\rightarrow {\mathbb R}$ as in ( H ). Following Theorems 1.16 and 1.18, item (d) yields a point of continuity $y\in [0,1]$ of h. If $\frac {1}{2^{m_{0}+1}}=h(y)>0$ , then by definition $y\in X_{m_{0}}$ where $m_{0}\in {\mathbb N}$ is the least such number. Since h is continuous at y, there is $N\in {\mathbb N}$ such that $z\in B(y, \frac {1}{2^{N}})$ implies $|h(z)-h(y)|<\frac {1}{2^{m_{0}+2}}$ . However, $O_{m_{0}+2}$ is dense in $[0,1]$ , implying there is $z_{0}\in B(y, \frac {1}{2^{N}})\cap O_{m_{0}+2}$ . Since $h(z_{0})\leq \frac {1}{2^{m_{0}+3}}$ by definition, we obtain a contradiction. Hence, we must have $h(y)=0$ and the latter implies $y\in \cap _{n\in {\mathbb N}}O_{n}$ by definition, as required for ${\textsf {BCT}}_{[0,1]}$ .

Secondly, for item (b), we first formulate a proof (say in ${\textsf {Z}}_{2}^{\Omega }$ ) and then show how this proof goes through in ${\textsf {ACA}}_{0}^{\omega }+{\textsf {BCT}}_{[0,1]}$ . Thus, fix lsco $f:[0,1]\rightarrow {\mathbb R}$ and define $C_{q}:=\{x\in [0,1]: f(x)\leq q\}$ , which is closed by definition. Define $F_{q}$ as $C_{q}\setminus \textsf {int}(C_{q})$ , where the open set $\textsf {int}(X)$ is the interior of X, i.e., those $x\in X$ such that there is $N\in {\mathbb N}$ with $B(x, \frac {1}{2^{N}})\subset X$ . By definition, $F_{q}$ is closed and is nowhere dense, i.e., the Baire category theorem shows that there is $y\in \big ([0,1 ]\setminus \cup _{q\in {\mathbb Q}}F_{q}\big )$ . The Baire category theorem is provable in ${\textsf {Z}}_{2}^{\Omega }$ by [Reference Normann and Sanders71, Section 6].

We now show that for each $x\in [0,1]$ where f is discontinuous, there is $q\in {\mathbb Q}$ such that $x\in F_{q}$ , establishing that f is continuous at the aforementioned point y, as required. Thus, let f be discontinuous at $x_{0}\in [0,1]$ and note that f cannot be usco at $x_{0}\in [0,1]$ , i.e., we have

(2.1) $$ \begin{align}\textstyle (\exists l\in {\mathbb N})(\forall N\in {\mathbb N})(\exists z\in B(x_{0}, \frac{1}{2^{N}}))(f(z)\geq f(x_{0})+\frac{1}{2^{l}}). \end{align} $$

Let $l_{0}$ be as in (2.1) and consider $q_{0}\in {\mathbb Q}$ such that $f(x_{0})<q_{0}< f(x_{0})+\frac {1}{2^{l_{0}}}$ . By definition, $x_{0}\in F_{q_{0}}$ , as required.

We now show how the above proof goes through in ${\textsf {ACA}}_{0}^{\omega }+{\textsf {BCT}}_{[0,1]}$ . The (only) two obstacles are as follows:

• • The definition of $\textsf {int}(X)$ involves quantification over ${\mathbb R}$ , i.e., the former set cannot be defined in ${\textsf {ACA}}_{0}^{\omega }$ .

• • The formula (2.1) involves quantification over ${\mathbb R}$ , i.e., the numbers $l_{0}$ and $q_{0}$ cannot be defined in terms of $x_{0}$ .

Both problems have (about) the same solution, as follows. For the second item, since f is lsco, (2.1) is equivalent to

(2.2) $$ \begin{align}\textstyle (\exists l\in {\mathbb N})(\forall N\in {\mathbb N})\underline{(\exists r\in B(x, \frac{1}{2^{N}})\cap {\mathbb Q})}(f(r)\geq f(x_{0})+\frac{1}{2^{l}}), \end{align} $$

where the underlined quantifier is crucial. Since (2.2) is decidable using $\exists ^{2}$ , the functional $\mu ^{2}$ computes $l_{1}\in {\mathbb N}$ , the least $l\in {\mathbb N}$ satisfying (2.2); clearly, $l=l_{1}+1$ then witnesses (2.1). Hence, $q_{0}$ from the previous paragraph is definable in terms of $x_{0}$ (and f and $\mu ^{2}$ ).

For the first item, i.e., relating to the interior of sets, the following formula expresses that $x\in F_{q}= (C_{q}\setminus \textsf {int}(C_{q})$ , by definition:

(2.3) $$ \begin{align}\textstyle f(x)\leq q \wedge (\forall N\in {\mathbb N})(\exists z\in B(x, \frac{1}{2^{N}}))( f(z)>q ). \end{align} $$

Similar to the second item and (2.1), since f is lsco, (2.3) is equivalent to

(2.4) $$ \begin{align}\textstyle f(x)\leq q \wedge (\forall N\in {\mathbb N})\underline{(\exists r\in B(x, \frac{1}{2^{N}})\cap {\mathbb Q})}( f(r)>q ), \end{align} $$

where again the underlined quantifier is crucial. Hence, we can define $F_{q}$ using $\exists ^{2}$ and (2.4). Using an enumeration of ${\mathbb Q}$ , one readily obtains an increasing sequence of closed nowhere dense sets. The rest of the proof now goes through. Since for lsco $f:[0,1]\rightarrow {\mathbb R}$ , the function $-f$ is usco, this item is finished.

For item (e), it is clear that h from ( H ) is countably continuous as it is constant on each $X_{n}$ and the complement of $\cup _{n\in {\mathbb N}}X_{n}$ .

To prove item (f) from ${\textsf {BCT}}_{[0,1]}$ , fix usco $f:[0,1]\rightarrow {\mathbb R}$ and note that $C_{n}:= \{ x\in [0,1]: f(x)\geq -n\}$ is closed and satisfies $[0,1]=\cup _{n\in {\mathbb N}}C_{n}$ . By ${\textsf {BCT}}_{[0,1]}$ , at least one $C_{n}$ is not nowhere dense, as required for item (f). Now assume the latter and let $(X_{n})_{n\in {\mathbb N}}$ be an increasing sequence of closed sets such that $[0,1]=\cup _{n\in {\mathbb N}}X_{n}$ . Define $\tilde {h}:[0,1]\rightarrow {\mathbb R}$ as follows: $\tilde {h}(x)=-n$ in case $x\in X_{n}$ and $n\in {\mathbb N}$ is the least such number. Similar to $h:[0,1]\rightarrow {\mathbb R}$ from ( H ), the function $\tilde {h}$ is usco. Item (f) provides $N\in {\mathbb N}$ and $c, d\in [0,1]$ such that $\tilde {h}(y)\geq -N$ for $y\in (c, d)$ . By definition, we have $(c, d)\subset X_{N}$ , i.e., the latter is not nowhere dense. Item (g) follows in the same way using h from ( H ) and $C_{n}:=\{x\in [0,1]: f(x)\geq \frac {1}{2^{n}} \}$ .

Proof First of all, it is a matter of definitions to show the equivalence between:

• • $g:{\mathbb R}\rightarrow {\mathbb R}$ is continuous at $x\in {\mathbb R}$ ,

• • $g:{\mathbb R}\rightarrow {\mathbb R}$ is usco and lsco at $x\in {\mathbb R}$ .

Secondly, for lsco $f:[0,1]\rightarrow {\mathbb R}$ , ‘f is discontinuous at $x\in [0,1]$ ’ is equivalent to

(2.5) $$ \begin{align}\textstyle (\exists l\in {\mathbb N})(\forall k\in {\mathbb N})(\exists y\in B(x, \frac{1}{2^{k}})(f(y)\geq f(x)+\frac{1}{2^{l}} ), \end{align} $$

which expresses that f is not usco at $x\in [0,1]$ . Now, (2.5) is equivalent to

(2.6) $$ \begin{align}\textstyle (\exists l\in {\mathbb N})(\forall k\in {\mathbb N})\underline{(\exists r\in B(x, \frac{1}{2^{k}})\cap {\mathbb Q})}(f(r)\geq f(x)+\frac{1}{2^{l}} ), \end{align} $$

where in particular the underlined quantifier in (2.6) has rational range due to f being lsco. Since (2.6) is arithmetical, $\exists ^{2}$ allows us to define $D_{f}$ , and hence $C_{f}$ .

Proof For the first item, we only need to derive ${\textsf {BCT}}_{[0,1]}$ from items (b)–(d) in the theorem with ‘continuity’ weakened to the notions from the corollary. This is a straightforward verification and we only establish the corollary for the weakest notion, namely lower quasi-continuity. Thus, assume item (b) with ‘lqco’ in the consequent and consider the real $y\in [0,1]$ as in the first paragraph of the proof of the theorem. If $\frac {1}{2^{m_{0}+1}}=h(y)>0$ , then by definition $y\in X_{m_{0}}$ where $m_{0}\in {\mathbb N}$ is the least such number. Since h is lqco at y, we have that for $\varepsilon =\frac {1}{2^{m_{0}+2}}$ and any $N\in {\mathbb N}$ , there is $(a, b)\subset B(y, \frac {1}{2^{N}})$ such that $z\in (a, b)$ implies $h(z)>h(y)-\frac {1}{2^{m_{0}+2}}=\frac {1}{2^{m_{0}+2}}$ . However, $O_{m_{0}+4}$ is dense in $[0,1]$ , implying there is $z_{0}\in (a, b)\cap O_{m_{0}+4}$ . Since $h(z_{0})\leq \frac {1}{2^{m_{0}+5}}$ , we obtain a contradiction. Hence, we must have $h(y)=0$ and the latter implies $y\in \cap _{n\in {\mathbb N}}O_{n}$ by definition, as required for ${\textsf {BCT}}_{[0,1]}$ .

For the second item, consider the first paragraph of the proof of Theorem 2.3. Assuming $[0,1]=\cup _{n\in {\mathbb N}}X_{n}$ , the usco function h from ( H ) is such that $C_{f}=\emptyset $ , i.e., definitely measure zero. Now proceed with the original proof and derive a contradiction. Thus, we may restrict the items (b)–(d) from Theorem 2.3 to usco f such that $C_{f}$ has measure zero.

For the third item, the Cantor set can be defined in ${\textsf {RCA}}_{0}$ (see [Reference Simpson93, p. 129]), and similar for ‘fat’ Cantor sets, i.e., closed nowhere dense sub-sets of $[0,1]$ that can have measure $1-\varepsilon $ for any $\varepsilon>0$ . Thus, let $(C_{n})_{n\in {\mathbb N}}$ be a sequence of fat Cantor sets in $[0,1]$ such that $C_{n}$ has measure at least $1-\frac {1}{2^{n}}$ for all $n\in {\mathbb N}$ . Now let $(O_{n})_{n\in {\mathbb N}}$ be any sequence of dense open sets in $[0,1]$ and note that $O_{n}\cap V_{n}$ is also open and dense, where $V_{n}=:[0,1]\setminus C_{n}$ . Since $\cap _{n\in {\mathbb N}} V_{n}$ has measure zero, the third item in the theorem applies to $(O_{n}\cap V_{n})_{n\in {\mathbb N}}$ , i.e., there is $y\in \cap _{n\in {\mathbb N}}\big (O_{n}\cap V_{n}\big )\subset \cap _{n\in {\mathbb N}}O_{n}$ , as required for ${\textsf {BCT}}_{[0,1]}$ .

Proof By definition, for usco $f:[0,1]\rightarrow {\mathbb R}$ , the set $\{x\in [0,1]: f(x)\geq a\}$ is closed for any $a\in {\mathbb R}$ . Let $O_{a}$ be the complement of the previous set. When given a modulus $\Psi $ , one readily defines $Y:[0,1]\rightarrow {\mathbb R}$ such that $B(y, Y(y))\subset O_{a}$ in case $y\in O_{a}$ . Hence, $O_{a}$ is R2-open and the Baire category theorem for such sets is provable in ${\textsf {ACA}}_{0}^{\omega }$ by [Reference Normann and Sanders71, Theorem 7.10]. One can now repeat the (modified) proof of item (b) in Theorem 2.3 in ${\textsf {ACA}}_{0}^{\omega }$ , noting that $\textsf {int}(C_{q})$ has an R2-open representation in light of the equivalence between (2.5) and (2.6).

Proof We use the proof of (a) $\rightarrow $ (b) from Theorem 2.3. In this proof, one defines a sequence of closed and nowhere dense sets $(F_{q})_{q\in {\mathbb Q}}$ such that the union contains $D_{f}$ for lsco $f:[0,1]\rightarrow {\mathbb R}$ . Applying ${\textsf {BCT}}_{[0,1]}$ , the set $C_{f}$ is observed to be non-empty. We show that the extra information provided by the Baire 1 representation of f allows us to define an R2-representation for $F_{q}$ . Instead of ${\textsf {BCT}}_{[0,1]}$ , we can then apply the Baire category theorem for R2-representations, which is provable in ${\textsf {ACA}}_{0}^{\omega }$ by [Reference Normann and Sanders71, Theorem 7.10]. In particular, we note that by (the proof of) [Reference Normann and Sanders79, Theorem 2.9], there is $\Phi :({\mathbb R}\times {\mathbb N})\rightarrow {\mathbb R}$ such that $\Phi (x, N)$ equals $\inf _{y\in B(x, \frac {1}{2^{N}})}f(y)$ for Baire 1 f.

Let $f:[0,1]\rightarrow {\mathbb R}$ be lsco and let $(f_{n})_{n\in {\mathbb N}}$ be a sequence of continuous functions with pointwise limit f on the unit interval. Recall the set $F_{q}\subset [0,1]$ defined as those $x\in [0,1]$ satisfying (2.4), where the latter is equivalent to (2.3). Define $O_{q}:=[0,1]\setminus F_{q}$ and note that ‘ $x\in O_{q}$ ’ is exactly

(2.7) $$ \begin{align}\textstyle f(x)> q \vee (\exists N\in {\mathbb N}){(\forall r\in B(x, \frac{1}{2^{N}})\cap {\mathbb Q})}( f(r)\leq q ). \end{align} $$

In case $x\in O_{q}$ satisfies the second disjunct in (2.7), we use $\mu ^{2}$ to find the least $N\in {\mathbb N}$ such that $\big (B(x, \frac {1}{2^{N}})\cap {\mathbb Q}\big )\subset O_{q}$ . Now, since f is lsco, the latter implies $B(x, \frac {1}{2^{N}})\subset O_{q}$ , as required for a R2-representation. In case $x\in O_{q}$ satisfies the first disjunct in (2.7), observe that

(2.8) $$ \begin{align}\textstyle f(x)>q \leftrightarrow (\exists N\in {\mathbb N})(q<\inf_{y\in B(x, \frac{1}{2^{N}})}f(y) ), \end{align} $$

as f is lsco. Since the infimum in the right-hand side of (2.8) is given by a function $\Phi :({\mathbb R}\times {\mathbb N})\rightarrow {\mathbb R}$ , we may use $\mu ^{2}$ to find the least $N_{0}\in {\mathbb N}$ such that $q<\inf _{y\in B(x, \frac {1}{2^{N_{0}}})}f(y)$ . By the definition of $O_{q}$ , we have $B(x, \frac {1}{2^{N_{0}}})\subset O_{q}$ , as required for a R2-representation. Hence, we have obtained a R2-representation of $O_{q}$ based on the case distinction (decidable by $\exists ^{2}$ ) provided by (2.7).

Proof The first part is immediate from Theorem 2.3 and the theorem. For the second part, a set $C\subset [0,1]$ is closed if and only if $\mathbb {1}_{C}$ is usco. Now, in case $x\in O:=[0,1]\setminus C$ , we have $(\exists N\in {\mathbb N})(0= \sup _{y\in B(x, \frac {1}{2^{N}})} \mathbb {1}_{C}(y) )$ . As noted in the proof of the theorem, the latter supremum is not just notation but given by a functional in case the function is also Baire 1. Hence, we can use $\mu ^{2}$ to define $Y:{\mathbb R}\rightarrow {\mathbb N}$ such that $B(x, \frac {1}{2^{Y(x)}})\subset O$ in case $x\in O$ . As $\mu ^{2}$ returns the least witness, $\cup _{q\in O\cap {\mathbb Q}}B(q, \frac {1}{2^{Y(q)+1}})$ is an RM-code for O, and we are done.

Proof The supremum principle for bounded effectively Baire n ( $n\geq 2$ ) functions is provable in ${\textsf {ACA}}_{0}^{\omega }+\Pi _{1}^{1}\text {-}{\textsf {CA}}_{0}$ by [Reference Normann and Sanders79, Theorem 2.22].

Proof Let $(f_{n})_{n\in {\mathbb N}}$ be a sequence of continuous functions with pointwise limit f. By [Reference Normann and Sanders79]Cor. 2.5, there is an associated sequence of RM-codes $(\Phi _{n})_{n\in {\mathbb N}}$ . For the second item, define the closed set $C_{n}$ as follows:

$$\begin{align*}C_{n}:= \{ x\in [0,1]: (\forall m\in {\mathbb N})(|f_{m}(x)|\leq n ) \}. \end{align*}$$

Thanks to the RM-codes $\Phi _{n}$ for $f_{n}$ , the formula $(\forall m\in {\mathbb N})(f_{m}(x)\leq n )$ is (second-order) $\Pi _{1}^{0}$ . Hence, the sets $C_{n}$ have corresponding RM-codes by [Reference Simpson93, II.5.7]. Since $[0,1]=\cup _{n\in {\mathbb N}}C_{n}$ , the Baire category theorem for RM-codes of open sets (provable in ${\textsf {RCA}}_{0}^{\omega }$ by [Reference Simpson93, II.5.8]), yields that at least one $C_{n}$ is not nowhere dense.

For the first item, define $D_{m,n}:=\{ x\in [0,1]: (\forall k\geq m)(|f_{k}(x)-f_{m}(x)|\leq \frac {1}{2^{n}} ) $ . As in the previous paragraph, each $D_{m,n}$ is RM-closed. To define the interior of the latter sets, put $x\in \textsf {int}(D_{m,n} )$ if and only if $(\exists N\in {\mathbb N} )(\forall r \in B(x, \frac {1}{2^{N}})\cap {\mathbb Q})(r \in D_{m,n})$ . Due to the continuity of the $f_{n}$ , $x\in \textsf {int}(D_{m,n} )$ is equivalent to

$$\begin{align*}\textstyle (\exists N\in {\mathbb N} )(\forall y \in B(x, \frac{1}{2^{N}}))(y \in D_{m,n}), \end{align*}$$

i.e., the ‘real’ definition of the interior of $D_{m,n}$ As in the proof of Theorem 2.3, define $F_{m,n}:=D_{m,n}\setminus \textsf {int}(D_{m,n} )$ , which is RM-closed and nowhere dense, and note that each point where f is discontinuous, is included in $\cup _{m,n\in {\mathbb N}}F_{m,n}$ . The Baire category theorem for RM-codes (provable in ${\textsf {RCA}}_{0}$ by [Reference Simpson93, II.5.7]) finishes the proof.

Finally, for the third item, as noted in the proof of Theorem 2.7, we have access to the supremum operator for Baire 1 functions following [Reference Normann and Sanders79, Theorem 2.9]. Hence, one readily defines the oscillation function using $\exists ^{2}$ .

Proof The proof of [Reference Menkyna60, Lemma 5] goes through with no modification since we assume an upper and lower bound for f.

Proof Assume ${\textsf {BCT}}_{[0,1]}$ and note that Theorem 2.3 implies that for usco $g_{0}$ and lsco $g_{1}$ , the sets $C_{g_{0}}$ and $C_{g_{1}}$ are dense, and hence the intersection is non-empty. Theorem 2.11 now implies items (b) and (c).

Now assume item (c) and suppose $(X_{n})_{n\in {\mathbb N}}$ is an increasing sequence of nowhere dense sets such that $[0,1]=\cup _{n\in {\mathbb N}}X_{n}$ . Now consider $h:[0,1]\rightarrow {\mathbb R}$ from ( H ), which is Baire $1^{*}$ as $f_{\upharpoonright X_{n}}$ is continuous for any $n\in {\mathbb N}$ . Item (c) provides a point of continuity, which yields $x\not \in \cup _{n\in {\mathbb N}}X_{n}$ as in the proof of Theorem 2.3, as required.

Proof The decomposition $f=g_{0}+g_{1}$ from Theorem 2.11 is provided by the proof of [Reference Menkyna60, Lemma 5]. By the latter, a Baire 1 representation of f also provides a Baire 1 representation of $g_{0}$ and $g_{1}$ . Now apply Theorem 2.7 to conclude that $C_{g_{0}}$ and $C_{g_{1}}$ are non-empty (and dense). Then $C_{g_{0}}\cap C_{g_{1}}\ne \emptyset $ , and we are done.

Proof Assume ${\textsf {BCT}}_{[0,1]}$ and let $(f_{n})_{n\in {\mathbb N}}$ be a sequence of lsco functions that is pointwise bounded on $[0,1]$ . By definition, the set $E_{n,N}:=\{x\in [0,1]: f_{n}(x)\leq N\}$ is closed, and so is $F_{N}:=\cap _{n\in {\mathbb N}}E_{n, N}$ . By pointwise boundedness, we have $[0,1]=\cup _{N\in {\mathbb N}}F_{N}$ and ${\textsf {BCT}}_{[0,1]}$ implies there is $N_{0}\in {\mathbb N}$ such that $(c, d)\subset F_{N_{0}}$ for some $c, d\in [0,1]$ . This interval is as required for uniform boundedness in item (b).

For the reversal, assume item (b) and define $\hat {h}:[0,1]\rightarrow {\mathbb R}$ as follows:

(2.9) $$ \begin{align} \hat{h}(x):= \begin{cases} 0, & x\not\in \cup_{n\in {\mathbb N}}X_{n},\\ n, & x\in X_{n} \text{ and } n\in {\mathbb N} \text{ is the least such number,} \end{cases} \end{align} $$

where $(X_{n})_{n\in {\mathbb N}}$ is an increasing sequence of closed sets such that $[0,1]=\cup _{n\in {\mathbb N}}X_{n}$ . Now define $\hat {h}_{n}:[0,1]\rightarrow {\mathbb R}$ as follows: $\hat {h}_{n}(x):=\hat {h}(x)$ in case $\hat {h}(x)<n$ , and $\hat {h}_{n}(x):=n$ otherwise. One readily verifies that the functions $\hat {h}, \hat {h}_{n}$ are lsco using the definition. Now, $(\hat {h}_{n})_{n\in {\mathbb N}}$ is pointwise bounded on $[0,1]$ , as $(\forall x\in X)(\forall n\in {\mathbb N})(\hat {h}_{n}(x)\leq \hat {h}(x) )$ by definition. Apply item (b) to find $(c, d)\subset [0,1]$ where $(\hat {h}_{n})_{n\in {\mathbb N}}$ is uniformly bounded, i.e., $(\exists M\in {\mathbb N})(\forall m\in {\mathbb N}, x\in (c, d))(\hat {h}_{m}(x)\leq M )$ . Taking $m=M+1$ , we note that $(c, d)\subset X_{M}$ , as required for ${\textsf {BCT}}_{[0,1]}$ .

Proof First of all, assume item (b) and fix a decreasing sequence of dense open sets $(O_{n})_{n\in {\mathbb N}}$ . Define the nowhere dense closed set $X_{n}:= [0,1]\setminus O_{n}$ and consider $h:[0,1]\rightarrow {\mathbb R}$ as in ( H ). Following Theorem 1.18, item (b) yields a point of continuity $y\in [0,1]$ of h. If $\frac {1}{2^{m_{0}+1}}=h(y)>0$ , then by definition $y\in X_{m_{0}}$ where $m_{0}\in {\mathbb N}$ is the least such number. Since h is continuous at y, there is $N\in {\mathbb N}$ such that $z\in B(y, \frac {1}{2^{N}})$ implies $|h(z)-h(y)|<\frac {1}{2^{m_{0}+2}}$ . However, $O_{m_{0}+4}$ is dense in $[0,1]$ , implying there is $z_{0}\in B(y, \frac {1}{2^{N}})\cap O_{m_{0}+3}$ . Since $h(z_{0})\leq \frac {1}{2^{m_{0}+5}}$ , we obtain a contradiction. Hence, we must have $h(y)=0$ and the latter implies $y\in \cap _{n\in {\mathbb N}}O_{n}$ by definition, as required for ${\textsf {BCT}}_{[0,1]}$ . As h from ( H ) is usco by Theorem 1.18, item (d) also applies. For item (c), one uses the proof of Theorem 2.5.

Secondly, assume ${\textsf {BCT}}_{[0,1]}$ and define $D_{m}:= \{ x\in [0,1]:{\textsf {osc}}_{f}(x)\geq \frac {1}{2^{m}} \}$ , which is closed by definition (or by Theorem 1.17). Suppose $D_{f}=\cup _{n\in {\mathbb N}}D_{n}$ equals $[0,1]$ . By ${\textsf {BCT}}_{[0,1]}$ , there is $n_{0}\in {\mathbb N}$ such that $D_{n_{0}}$ is not nowhere dense, i.e., there is $[a, b]\subset D_{n_{0}} $ . Now fix $x_{0}\in [0,1]$ , $\varepsilon _{0}:=\frac {1}{2^{n_{0}+1}}$ , and $N_{0}\in {\mathbb N}$ such that $B(x_{0},\frac {1}{2^{N_{0}}})\subset (a, b)$ . Since f is cliquish, there is an interval $(c, d)\subset B(x_{0},\frac {1}{2^{N_{0}}}) $ such that for all $z, w\in (c, d)$ , we have $|f(z)-f(w)|<\varepsilon _{0}$ . The latter implies that ${\textsf {osc}}_{f}(z)\leq \varepsilon _{0}$ for $z \in (c, d)$ , which, however, contradicts the fact that $z\in (c, d)\subset D_{n_{0}}$ . Hence, $D_{f}$ does not equal $[0,1]$ , as required for item (b).

Fourth, item (d) follows since h from ( H ) is usco by Theorem 1.18.

Fifth, let $f:[0,1]\rightarrow {\mathbb R}$ be as in item (e) and define the closed set $D_{k}:=\{ x\in [0,1]: {\textsf {osc}}_{f}(x)\geq \frac {1}{2^{k}}\}$ . By assumption, $[0,1]=\cup _{k\in {\mathbb N}}D_{k}$ and ${\textsf {BCT}}_{[0,1]}$ implies there is $k_{0}\in {\mathbb N}$ and $(c, d)\subset [0,1]$ such that $(c, d)\subset D_{k_{0}}$ , as required. For the reversal, assume item (e) and let $f:[0,1]\rightarrow {\mathbb R}$ be cliquish with oscillation function ${\textsf {osc}}_{f}:[0,1]\rightarrow {\mathbb R}$ . In case $C_{f}\ne \emptyset $ , we are done in light of item (b). In case $C_{f}=\emptyset $ , f is totally discontinuous and we can apply item (e). Let $N_{0}\in {\mathbb N}$ and $c, d\in [0,1]$ be such that ${\textsf {osc}}_{f}(x)\geq \frac {1}{2^{N_{0}}}$ for $x\in (c, d)$ . Now fix $\varepsilon _{0}>0$ and $x_{0}\in (c, d)$ such that $ \varepsilon _{0}<\frac {1}{2^{N_{0}}}$ and $B(x_{0}, \varepsilon _{0})\subset (c,d)$ ; as f is cliquish, there is an interval $(a, b)\subset B(x_{0}, \varepsilon _{0})$ such that for all $x, y\in (a, b)$ , we have $|f(x)-f(y)|<\varepsilon _{0}$ . However, this implies that ${\textsf {osc}}_{f}(z)\leq \frac {1}{2^{N_{0}}}$ for $z\in (a, b)$ , contradicting ${\textsf {osc}}_{f}(z)>\frac {1}{2^{N_{0}}}$ by $z\in (a, b)\subset (c, d)$ .

Proof The first item implies the second item by Theorem 2.16. Now assume the second item and let $(X_{n})_{n\in {\mathbb N}}$ be an increasing sequence of closed nowhere dense sets such that $[0,1]=\cup _{n\in {\mathbb N}}X_{n}$ . Define $Z(x)$ as the least $n\in {\mathbb N}$ such that $x\in X_{n}$ and $e(x):= \sum _{n=0}^{Z(x)+1}\frac {x^{n}}{n!} $ , implying $e(x)<e^{x}$ for $x\in [0,1]$ and $\sup _{x\in [p,q]}e(x)=e^{q}$ by definition. To show that $\lambda x.e(x)$ is cliquish, fix $x_{0}\in [0,1]$ and consider

$$\begin{align*}|e(y)-e(z)|\leq |e(y)-e^{y}|+|e^{y}-e^{z}|+|e(z)-e^{z}|, \end{align*}$$

to which the usual ‘epsilon over three’ trick applies as follows: the middle term can be made arbitrarily small thanks to the uniform continuity of $\lambda x.e^{x}$ on $[0,1]$ . The other two terms can be made arbitrarily small by picking large enough $k\in {\mathbb N}$ and picking $x_{1}\in O_{k}:=[0,1]\setminus X_{k}$ close enough to $x_{0}$ (using the density of the sets $O_{n}$ ). Indeed, $B(x_{1}, \frac {1}{2^{N_{0}}})\subset O_{k}$ for some $N_{0}\in {\mathbb N}$ , implying that $Y(w)\geq k$ for w in the former ball, i.e., $e(w)$ approaches $e^{w}$ as k grows. Since $\lambda x.e(x)$ is totally discontinuous, item (b) yields a contradiction. Hence, there must be $y\not \in \cup _{n\in {\mathbb N}}X_{n}$ as required for ${\textsf {BCT}}_{[0,1]}$ , and we are done.

Proof First of all, sico functions are cliquish (say in ${\textsf {ZFC}}$ ) while the proof of [Reference Neubrunnová67, Theorem 2.1] immediately establishes that sequentially sico functions are cliquish, working over ${\textsf {ACA}}_{0}^{\omega }+{\textsf {BCT}}_{[0,1]}$ . Hence, item (b) immediately follows from ${\textsf {BCT}}_{[0,1]}$ via Theorem 2.16.

Secondly, let $(X_{n})_{n\in {\mathbb N}}$ be an increasing sequence of closed and nowhere dense sets such that $[0,1]=\cup _{n\in {\mathbb N}}X_{n}$ and consider $h:[0,1]\rightarrow {\mathbb R}$ as in ( H ). Fix $G\subset {\mathbb R}$ and consider the following two cases which show that h is (sequentially) sico.

• • In case $(\forall n\in {\mathbb N})(\exists m> n)( \frac {1}{2^{m+1}}\in G)$ , we have $h^{-1}(G)=\cup _{n\in {\mathbb N}}X_{n}=[0,1]$ .

• • In case $n_{0}\in {\mathbb N}$ is the least n with $(\forall m> n)( \frac {1}{2^{m+1}}\not \in G)$ , then $h^{-1}(G)= X_{n_{0}}$ .

Note that $\mu ^{2}$ can find the number $n_{0}$ from the second item and decide which case holds, i.e., h is also sequentially sico. Following Theorem 1.16, item (b) now provides a point of continuity for h, and ${\textsf {BCT}}_{[0,1]}$ readily follows.

Proof Fix quasi-continuous $f:[0,1]\rightarrow {\mathbb R}$ and use $\exists ^{2}$ to define $x\in O_{m}$ in case

(2.10) $$ \begin{align}\textstyle (\exists N\in {\mathbb N})(\forall q, r\in B(x, \frac{1}{2^{N}})\cap {\mathbb Q})( |f(q)-f(r)|\leq \frac{1}{2^{m}} ). \end{align} $$

By quasi-continuity, the formula (2.10) is equivalent to

(2.11) $$ \begin{align}\textstyle (\exists M\in {\mathbb N})(\forall w, z\in B(x, \frac{1}{2^{M}}))( |f(w)-f(z)|\leq \frac{1}{2^{m}} ), \end{align} $$

which immediately implies that $O_{m}$ is open. Apply $\mu ^{2}$ to (2.10) to obtain an R2-representation of $O_{m}$ . Now define $D_{m}:= [0,1]\setminus O_{m}$ and note that $D_{f}:=\cup _{m\in {\mathbb N}}D_{m}$ contains all points where f is discontinuous (essentially by definition). By [Reference Normann and Sanders71, Theorem 7.10], ${\textsf {ACA}}_{0}^{\omega }$ proves ${\textsf {BCT}}_{[0,1]}$ for R2-open sets, i.e., the first and second items now follow. Finally, by [Reference Normann and Sanders79, Theorem 2.9], the supremum and infimum of f exist (uniformly on any interval) given $\exists ^{2}$ . Hence, the oscillation function as in the third item is readily defined, where we note that $D_{m}$ is by definition $\{x\in [0,1]:{\textsf {osc}}_{f}(x)\geq \frac {1}{2^{m}}\}$ , i.e., the usual definition.

Principle 2.22 ( ${\textsf {IND}}_{{\mathbb R}}$ ).

For $F:({\mathbb R}\times {\mathbb N})\rightarrow {\mathbb N}, k\in {\mathbb N}$ , there is $X\subset {\mathbb N}$ such that

$$\begin{align*}(\forall n\leq k)\big[ (\exists x\in {\mathbb R})(F(x, n)=0)\leftrightarrow n\in X\big]. \end{align*}$$

Proof First of all, we show that fragmented functions are cliquish (in ${\textsf {ACA}}_{0}^{\omega }$ ); Theorem 2.16 then establishes the implication from ${\textsf {BCT}}_{[0,1]}$ to item (b). Let $f:[0,1]\rightarrow {\mathbb R}$ be fragmented and fix $\varepsilon>0$ , $N\in {\mathbb N}$ , and $x\in [0,1]$ . For closed $C:=[x_{0}-\frac {1}{2^{N_{0}}}, x_{0}+\frac {1}{2^{N_{0}}}]$ , there is a relatively open $O\subset C$ with ${\textsf {diam}}(f(O))<\varepsilon $ . Since C is a closed interval, the relative openness of O implies that there are $a, b\in {\mathbb R}$ with $(a, b)\subset O$ . Since ${\textsf {diam}}(f(O))<\varepsilon $ , we have $|f(x)-f(y)|<\varepsilon $ for $x, y\in (a, b)\subset B(x, \frac {1}{2^{N_{0}}})$ , i.e., f is cliquish.

Secondly, we show that item (b) implies ${\textsf {BCT}}_{[0,1]}$ . To this end, let $(X_{n})_{n\in {\mathbb N}}$ be an increasing sequence of closed nowhere dense sets in $[0,1]$ such that $[0,1]=\cup _{n\in {\mathbb N}}X_{n}$ . We now show that $h:[0,1]\rightarrow {\mathbb R}$ from ( H ) is fragmented. Fix $\varepsilon>0$ and closed $C\subset [0,1]$ and consider the following case distinction.

• • In case $C\subset X_{0}$ , put $O:= C$ and note that O is relatively open (in C). Moreover, $f(x)=\frac {1}{2}$ for $x\in O\subset X_{0}$ , i.e., Definition 2.21 is trivially satisfied.

• • In case $C\subset \cup _{k\leq k_{0}} X_{k}$ for (minimal) $k_{0}>0$ , put $O:= C\cap O_{k_{0}-1}$ where $O_{k}:=[0,1]\setminus X_{k}$ . Since $O_{k_{0}-1}$ is open, O is relatively open (in C). Similar to the previous item, $f(x)=\frac {1}{2^{k_{0}}}$ for $x\in O$ , i.e., Definition 2.21 is again trivially satisfied.

• • For the remaining case, since $C\cap X_{k}\ne \emptyset $ for any $k\in {\mathbb N}$ , choose $k_{0}\in {\mathbb N}\setminus \{0\}$ with $\frac {1}{2^{k_{0}}}<\varepsilon $ and define $O:= C\cap O_{k_{0}-1}$ . Similar to the previous item, $O_{k_{0}-1}$ is open and O is relatively open (in C). By assumption, $f(x)< \frac {1}{2^{k_{0}}}$ for $x\in O\subset [0,1]\setminus X_{k_{0}}$ . Hence, ${\textsf {diam}}(f(O))\leq \frac {1}{2^{k_{0}}}<\varepsilon $ , as required.

Hence h from ( H ) is fragmented, while the second item seems to need ${\textsf {IND}}_{{\mathbb R}}$ to guarantee that $k_{0}$ is minimal. We also obtain an oscillation function by Theorem 1.16, i.e., we may apply item (b) to show that h is continuous at some $x_{0}\in [0,1]$ . As in the proof of Theorem 2.16, we obtain ${\textsf {BCT}}_{[0,1]}$ , and we are done.

Thirdly, item (c) implies ${\textsf {BCT}}_{[0,1]}$ as follows: let $(X_{n})_{n\in {\mathbb N}}$ be an increasing sequence of closed nowhere dense sets in $[0,1]$ such that $[0,1]=\cup _{n\in {\mathbb N}}X_{n}$ . Now consider h from ( H ) and note that for any $G\subset {\mathbb R}$ , $h^{-1}(G)$ is the union of all $X_{n}$ such that $\frac {1}{2^{n+1}}\in G$ , i.e., $\mathbf {F}_{\sigma }$ by definition. Together with Theorem 1.18, we observe that item (c) applies to h; as in the proof of Theorem 2.16, we obtain ${\textsf {BCT}}_{[0,1]}$ . The other direction is immediate by Theorem 2.16.

Proof The first part is immediate from Theorem 2.10 and the theorem. For the second part, one readily shows that for closed $C\subset [0,1]$ , the function $\mathbb {1}_{C}$ is fragmented. As in the proof of Corollary 2.8, one obtains an RM-code for C.

Proof First of all, to prove item (d) using ${\textsf {BCT}}_{[0,1]}$ , suppose $f, g:[0,1]\rightarrow {\mathbb R}$ are usco and such that $C_{f}=D_{g}$ and $C_{g}=D_{f}$ , noting that these sets exist by Theorem 2.4. Now consider the (closed and nowhere dense) sets $F_{q}$ from the proof of Theorem 2.3 and let $H_{q}$ be the associated (closed and nowhere dense) sets for g. Since $D_{f}\subset \cup _{q\in {\mathbb Q}}F_{q}$ and $D_{g}\subset \cup _{q\in {\mathbb Q}}H_{q} $ , we have $ [0,1]=C_{f}\cup D_{f}=D_{g}\cup D_{f}=\cup _{q\in {\mathbb Q}}(F_{q}\cup H_{q})$ , which contradicts ${\textsf {BCT}}_{[0,1]}$ .

A similar proof goes through for item (c), namely by defining $D_{k}^{f}:= \{x\in [0,1]: {\textsf {osc}}_{f}(x)\geq \frac {1}{2^{k}}\}$ and noting that $[0,1]=\cup _{n\in {\mathbb N}}(D_{n}^{f}\cup D_{n}^{g}) $ where the latter sets are nowhere dense as in the proof of Theorem 2.16. Items (e) and (f) follow from items (c) and (d) as Thomae’s function from ( T ) is continuous on the irrationals and discontinuous on ${\mathbb Q}$ ; this function is also usco and cliquish. By definition, pointwise discontinuous functions are cliquish, i.e., item (b) also follows.

Secondly, we now derive ${\textsf {BCT}}_{[0,1]}$ from Volterra’s corollary as in item (e) or (f). To this end, let $(X_{n})_{n\in {\mathbb N}}$ be a sequence of closed nowhere dense sets such that $[0,1]\setminus {\mathbb Q}=\cup _{n\in {\mathbb N}}X_{n}$ . Now consider $h:[0,1]\rightarrow {\mathbb R}$ as in ( H ), which is as required for item (e) or (f) by Theorems 1.16 and 1.18. By Theorem 1.16, h is also continuous at any $q\in {\mathbb Q}\cap [0,1]$ (since $h(q)={\textsf {osc}}_{h}(q)=0$ ) and discontinuous at any $y\in [0,1]\setminus {\mathbb Q}$ (since $h(y)={\textsf {osc}}_{h}(y)>0$ ). This contradicts Volterra’s corollary (for usco or cliquish functions), and ${\textsf {BCT}}_{[0,1]}$ follows. The same argument works for item (b) using Thomae’s function and $h:[0,1]\rightarrow {\mathbb R}$ as the latter is pointwise discontinuous, namely continuous at every rational, by assumption.

Thirdly, we only need to show that ${\textsf {BCT}}_{[0,1]}$ implies item (g), as ${\mathbb Q}$ is enumerable and dense. To this end, proceed as in the previous paragraph, replacing ${\mathbb Q}$ by a countable dense set $D\subset [0,1]$ wherever relevant. Note in particular that $D=\cup _{n\in {\mathbb N}} \{x\in D: Y(x)= n\}$ suffices to replace an enumeration of ${\mathbb Q}$ if $Y:[0,1]\rightarrow {\mathbb N}$ is injective on D.

Fourth, ${\textsf {BCT}}_{[0,1]}$ implies items (h) by Theorem 2.3 since $D=C_{f}$ is as required. To prove (h) $\rightarrow {\textsf {BCT}}_{[0,1]}$ , let $(X_{n})_{n\in {\mathbb N}}$ be a sequence of closed and nowhere dense sets in $[0,1]$ and consider $h:[0,1]\rightarrow {\mathbb R}$ from ( H ), which is usco and cliquish by Theorem 1.18. Let D be the dense set provided by item (h) and consider $y\in D$ . In case $h(y)\ne 0$ , say $h(y)\geq \frac {1}{2^{k_{0}}}$ , note that $O_{k_{0}}\cap D$ is dense in $[0,1]$ . Hence, for any $N\in {\mathbb N}$ there is $z\in B(x, \frac {1}{2^{N}})\cap D$ with $h(z)\leq \frac {1}{2^{k_{0}+1}}$ , i.e., $h_{\upharpoonright D}$ is not continuous (on D). This contradiction implies that $h(y)=0$ , meaning $y\in [0,1]\setminus \cup _{n\in {\mathbb N}}X_{n}$ , i.e., ${\textsf {BCT}}_{[0,1]}$ follows. The final sentence is immediate in light of Theorem 2.11.

Proof For item (a), for quasi-continuous $f, g:[0,1]\rightarrow {\mathbb R}$ , the sets $D_{f}$ and $D_{g}$ exist and can be expressed as unions $\cup _{n\in {\mathbb N}}D_{n}^{f}$ and $\cup _{m\in {\mathbb N}}D_{m}^{g}$ of R2-closed nowhere dense sets by Theorem 2.19. Hence, in case $C_{f}=D_{g}$ and $C_{g}=D_{f}$ , we have $[0,1]=C_{f}\cup D_{f}=D_{g}\cup D_{f}=\bigcup _{n, m\in {\mathbb N}} D_{n}^{f}\cup D_{m}^{g} $ . However, the Baire category theorem for R2-closed sets is provable in ${\textsf {ACA}}_{0}^{\omega }$ [Reference Normann and Sanders71, Theorem 7.10], a contradiction. The other items are proved in the same way.

Proof Suppose $f:[0,1]\rightarrow {\mathbb R}$ is Riemann integrable and unbounded, i.e., $(\forall n\in {\mathbb N})(\exists x\in [0,1])(|f(x)|>n)$ . Apply ${\textsf {QF-AC}}^{0,1}$ to obtain a sequence $(x_{n})_{n\in {\mathbb N}}$ with $|f(x_{n})|>n$ for $n\in {\mathbb N}$ . By sequential compactness (see [Reference Simpson93, III.2.2]), there is a convergent sub-sequence $(y_{n})_{n\in {\mathbb N}}$ , say with limit $y\in [0,1]$ . In particular, $f(y_{n})$ is arbitrarily large as $y_{n}$ approaches y. Thus, for any Riemann sum $\sum _{i=0}^{k} f(t_{i}) (x_{i+1}-x_{i}) $ , consider the point $t_{j}$ such that $y\in [x_{j}, x_{j+1}]$ and the latter contains infinitely many elements of the sequence $(y_{n})_{n\in {\mathbb N}}$ . Changing this $t_{j}$ to $y_{m}$ for large enough $m\in {\mathbb N}$ , the Riemann sum can be arbitrarily large, a contradiction.