Connecting real and hyperarithmetical analysis

We first derive Finite Choice from item (a) via a proof-by-contradiction. To this end, fix a sequence of non-empty finite sets of reals $(X_{n})_{n\in{\mathbb{N}}}$. Suppose there is no sequence $(x_{n})_{n\in{\mathbb{N}}}$ of reals such that $x_{n}\in X_{n}$ for all $n\in{\mathbb{N}}$. We now define

where $w^{1^{*}}$ is a finite sequence of reals of length $|w|$, readily coded using $(\exists^{2})$. We define the equivalence relation ‘$=_{M_{0}}$’ as follows: the relation $w=_{M_{0}}v$ holds if $|w|=_{0}|v|$, where $w,v\in M_{0}$. The metric $d_{0}:M_{0}^{2}\rightarrow{\mathbb{R}}$ is defined as $d_{0}(w,v)=|\frac{1}{2^{|v|}}-\frac{1}{2^{|w|}}|$ for any $w,v\in M_{0}$. We then have $d_{0}(v,w)=_{{\mathbb{R}}}0\leftrightarrow|v|=_{0}|w|\leftrightarrow v=_{M_{0}}w$ as required. We also have $0\leq d_{0}(v,w)=_{{\mathbb{R}}}d_{0}(w,v)$ for any $v,w\in M_{0}$, while for any $z\in M_{0}$ we observe:

by the triangle equality of the absolute value on the reals. Hence, $(M_{0},d_{0})$ is a metric space as in Definition 2.1.

To show that $(M_{0},d_{0})$ is sequentially compact, let $(w_{n})_{n\in{\mathbb{N}}}$ be a sequence in $M_{0}$ and consider the following case distinction. In case $(\forall n\in{\mathbb{N}})(|w_{n}|<m_{0})$ for some fixed $m_{0}\in{\mathbb{N}}$, then $(w_{n})_{n\in{\mathbb{N}}}$ contains at most $(m_{0}+1)!$ different elements. The pigeon hole principle now implies that at least one $w_{n_{0}}$ occurs infinitely often in $(w_{n})_{n\in{\mathbb{N}}}$, i.e. $(w_{n_{0}})_{n\in{\mathbb{N}}}$ is a convergent sub-sequence. In case $(\forall m\in{\mathbb{N}})(\exists n\in{\mathbb{N}})(|w_{n}|\geq m)$, the sequence $(w_{n})_{n\in{\mathbb{N}}}$ yields a sequence $(x_{n})_{n\in{\mathbb{N}}}$ such that $x_{n}\in X_{n}$ for all $n\in{\mathbb{N}}$, which is impossible by assumption. Hence, $(M_{0},d_{0})$ is a sequentially compact metric space.

Next, define $f:M_{0}\rightarrow{\mathbb{R}}$ as $f(w):=|w|$, which is clearly not bounded on $M_{0}$, which one shows using $\textup{{IND}}_{2}$. To show that $f$ is continuous at $w_{0}\in M_{0}$, consider the formula $|\frac{1}{2^{|w_{0}|}}-\frac{1}{2^{|v|}}|=d_{0}(v,w_{0})<\frac{1}{2^{N}}$; the latter is false for $N\geq|w_{0}|+2$ and any $v\neq_{M_{0}}w_{0}$. Hence, the following formula is vacuously true:

i.e. $f$ is continuous at $w_{0}\in M_{0}$, with a modulus of continuity given by $h(w,k):=\frac{1}{2^{|w|+k+2}}$. To see that $f$ is also topologically continuous, fix an open set $V\subset{\mathbb{R}}$ and fix $w_{0}\in f^{-1}(V)$. Then for $N_{0}:=|w_{0}|+2$, one verifies that $B_{d}^{M_{0}}(w_{0},\frac{1}{2^{N_{0}}})\subset f^{-1}(V)$, i.e. $f^{-1}(V)$ is open. Thus, $f:M_{0}\rightarrow{\mathbb{R}}$ is a continuous but unbounded function on a sequentially compact metric space $(M_{0},d_{0})$, contradicting item (a). Item (b) also implies Finite Choice as $f$ is not uniformly continuous. For item (c), $g:M_{0}\rightarrow{\mathbb{R}}$ defined as $g(w):=\frac{1}{2^{|w|}}$ is continuous in the same way as for $f$. However, using $\textup{{IND}}_{2}$, the infimum of $g$ on $M_{0}$ is $0$, but there is no $w\in M_{0}$ with $g(w)=_{{\mathbb{R}}}0$, by definition. Hence, item (c) also implies Finite Choice.

Now assume item (d) and suppose Finite Choice is again false; letting $(X_{n})_{n\in{\mathbb{N}}}$ and $(M_{0},d_{0})$ be as in the previous paragraph, we define $f_{n}:({\mathbb{N}}\times M_{0})\rightarrow{\mathbb{R}}$ as:

Clearly, $\lim_{n\rightarrow\infty}f_{n}(w)=f(w)$ and $f_{n}(w)\leq f_{n+1}(w)$ for $w\in M_{0}$; $f_{n}$ is continuous in the same way as for $f$. Item (d) implies that the convergence is uniform, i.e.

which yields a contradiction by letting $N_{1}\in{\mathbb{N}}$ be as in (2.4) for $k=1$ and choosing $w_{1}\in M$ of length $N_{1}+1$ using $\textup{{IND}}_{2}$. One derives Finite Choice from item (e) in the same way.

Next, regarding item (g), suppose Finite Choice is false and consider again $(M_{0},d_{0})$. Define the continuous function $f:M_{0}\rightarrow{\mathbb{R}}$ by $f(w)=q_{|w|}$ where $(q_{n})_{n\in{\mathbb{N}}}$ is an enumeration of the rationals without repetitions. Using $\textup{{IND}}_{2}$, we have $f(M_{0})={\mathbb{Q}}$ and the latter is not closed while $M_{0}$ is, contradicting item (g), and Finite Choice must hold. To obtain the latter from item (f), note that $(M_{0},d_{0})$ is infinite (using $\textup{{IND}}_{2}$) while all functions $f:M_{0}\rightarrow{\mathbb{R}}$ are continuous as (2.2) is vacuously true. Regarding item (j), assuming again that Finite Choice is false, the sequence $(f_{n})_{n\in{\mathbb{N}}}$ as in (2.3) is equicontinuous:

which (vacuously) holds in the same way as for (2.2). However, as for item (d), uniform convergence (of a sub-sequence) is false, i.e. item (j) also implies Finite Choice. For item (i), suppose Finite Choice is false, define $C_{n}:=\{w\in M_{0}:|w|>n+1\}$, and verify that this closed and non-empty set has diameter at most $\frac{1}{2^{n}}$, using $\textup{{IND}}_{2}$. Since $\cap_{n\in{\mathbb{N}}}C_{n}=\emptyset$, we obtain Finite Choice from item (i). For item (h), suppose Finite Choice is false, and define $C=\{w\in M_{0}:|w|\textup{ is odd}\}$ and $D=\{w\in M_{0}:|w|\textup{ is even}\}$. One readily verifies that $C\cap D=\emptyset$, $C,D$ are closed, and $d(C,D)=0$.

To establish the items in the theorem in $\textup{{ACA}}_{0}^{\omega}+\textup{{QF-AC}}^{0,1}$, the usual proof-by-contradiction goes through. A proof sketch of item (a) as follows: let $(M,d)$ be as in the latter and suppose $f:M\rightarrow{\mathbb{R}}$ is continuous and unbounded, i.e. $(\forall n\in{\mathbb{N}})(\exists x\in M)(|f(x)|>n)$. Since $M\subset{\mathbb{R}}$ and real numbers are represented by elements of Baire space, we may apply $\textup{{QF-AC}}^{0,1}$ to obtain $(x_{n})_{n\in{\mathbb{N}}}$ in $M$ such that $|f(x_{n})|>n$ for all $n\in{\mathbb{N}}$. Since $M$ is sequentially compact, $(x_{n})_{n\in{\mathbb{N}}}$ has a convergent sub-sequence, say with limit $y\in M$. Clearly, $f$ is not continuous at $y\in M$, a contradiction. To obtain (f), apply $\textup{{QF-AC}}^{0,1}$ to the statement that $M\subset[0,1]$ is infinite, yielding a sequence $(w_{n})_{n\in{\mathbb{N}}}$ in $M$. Now define $f:M\rightarrow{\mathbb{R}}$ as $f(x_{n})=n$ and $f(y)=0$ for $y\neq x_{m}$ for all $m\in{\mathbb{N}}$. Since $f$ is unbounded on $M$, it is discontinuous by item (a). Most other items are established using $\textup{{QF-AC}}^{0,1}$ in the same way.

We also sketch how $\textup{{QF-AC}}^{0,1}$ implies item (g). To this end, let $f,M$ be as in the closed map lemma and suppose $f(C)$ is not closed for closed $C\subset M$. Hence, there is $y_{0}\not\in f(C)$ such that $(\forall k\in{\mathbb{N}})(\exists y\in B(y_{0},\frac{1}{2^{k}}))(y\in f(C))$. By definition, the latter formula means $(\forall k\in{\mathbb{N}})(\exists x\in C)(|f(x)-y_{0}|<\frac{1}{2^{k}})$. Apply $\textup{{QF-AC}}^{0,1}$ to obtain a sequence $(x_{n})_{n\in{\mathbb{N}}}$ in $C$ with $(\forall k\in{\mathbb{N}})(|f(x_{k})-y_{0}|<\frac{1}{2^{k}})$. By sequential compactness, there is a convergent sub-sequence $(z_{n})_{n\in{\mathbb{N}}}$, say with limit $z$. Since $C$ is closed, we have $z\in C$ and since $f$ is continuous (and hence sequentially continuous) also $f(z)=y_{0}$. This contradicts $y_{0}\not\in f(C)$ and the closed map lemma therefore follows from $\textup{{QF-AC}}^{0,1}$. ∎

We first derive Finite Choice from item (a) via a proof-by-contradiction. To this end, fix a sequence of non-empty finite sets of reals $(X_{n})_{n\in{\mathbb{N}}}$. Suppose there is no sequence $(x_{n})_{n\in{\mathbb{N}}}$ of reals such that $x_{n}\in X_{n}$ for all $n\in{\mathbb{N}}$ and recall $M_{0}$ from (2.1). Now define $M_{1}=M_{0}\cup\{0_{M_{1}}\}$ where $0_{M_{1}}$ is a new symbol such that $w\neq_{M_{1}}0$ for $w\in M_{0}$ and ‘$=_{M_{1}}$’ is ‘$=_{M_{0}}$’ otherwise. Define $d_{1}:M_{1}^{2}\rightarrow{\mathbb{R}}$ as $d_{0}$ on $M_{0}$, as $d_{1}(w,0_{M_{1}}):=d(0_{M_{1}},w)=\frac{1}{2^{|w|}}$ for $w\in M_{0}$, and $d_{1}(0_{M_{1}},0_{M_{1}})=0$. Then $(M_{1},d_{1})$ is a metric, which is shown in the same way as for $(M_{0},d_{0})$.

To show that $(M_{1},d_{1})$ is countably-compact, let $(O_{n})_{n\in{\mathbb{N}}}$ be an open cover of $M_{1}$ and suppose $n_{1}\in{\mathbb{N}}$ is sucht hat $0_{M_{1}}\in O_{n_{1}}$. By definition, there is $N_{1}\in{\mathbb{N}}$ such that $B_{d_{1}}^{M_{1}}(0_{M_{1}},\frac{1}{2^{N_{1}}})\subset O_{n_{0}}$, i.e. $d(0_{M_{1}},w)=\frac{1}{2^{|w|}}<\frac{1}{2^{N_{1}}}$ implies $w\in O_{n_{1}}$ for $w\in M_{0}$. Now use $\textup{{IND}}_{2}$ to enumerate the finitely many $v\in M_{0}$ such that $|v|\leq N_{1}$. This finite sequence is covered by some $\cup_{n\leq n_{2}}O_{n}$, i.e. we have obtained a finite sub-covering of $M_{1}$, namely $\cup_{n\leq\max(n_{1},n_{2})}O_{n}$. Moreover, $(M_{1},d_{1})$ is sequentially compact, which can be proved via the same case distinction as for $(M_{0},d_{0})$ in the proof of Theorem 2.5.

Next, define $g:M_{1}\rightarrow{\mathbb{R}}$ as $g(w):=|w|$ for $w\in M_{0}$ and $g(0_{M_{1}})=_{{\mathbb{R}}}0$, which is clearly not bounded on $M_{1}$; this follows again via $\textup{{IND}}_{2}$. Then $g$ is continuous at $w_{0}\in M_{0}$ in the same way as $f$ from the proof of Theorem 2.5, namely since (2.2) is vacuously true. To show that $g$ is sequentially continuous at $0_{M_{1}}$, let $(w_{n})_{n\in{\mathbb{N}}}$ be a sequence converging to $0_{M_{1}}$. In case this sequence is eventually constant $0_{M_{1}}$, clearly $g(0_{M})=\lim_{n\rightarrow\infty}g(w_{n})$ as required. In case $(w_{n})_{n\in{\mathbb{N}}}$ is not eventually constant $0_{M_{1}}$, the convergence to $0_{M_{1}}$ in the $d_{1}$-metric implies that for any $n\in{\mathbb{N}}$, there is $m\geq n$ with $|w_{m}|>n$. Thus, $(w_{n})_{n\in{\mathbb{N}}}$ yields a sequence $(x_{n})_{n\in{\mathbb{N}}}$ such that $x_{n}\in X_{n}$ for all $n\in{\mathbb{N}}$, which contradicts our assumptions, i.e. this case cannot occur. As a result, $g:M_{1}\rightarrow{\mathbb{R}}$ is sequentially continuous. Since, it is also unbounded (thanks to $\textup{{IND}}_{2}$), we obtain a contradiction with item (a). Thus, (a) implies Finite Choice, and the same for item (b). To obtain Finite Choice from item (c), note that $(M_{1},d_{1})$ is infinite (using $\textup{{IND}}_{2}$) while all functions $f:M_{1}\rightarrow{\mathbb{R}}$ are sequentially continuous by the previous.

To show that $(M_{1},d_{1})$ satisfies the properties in item (d), note that for $\Psi:M_{1}\rightarrow{\mathbb{R}}^{+}$, the ball $B_{d_{1}}^{M_{1}}(0_{M_{1}},\Psi(0_{M_{1}}))$ covers all but finitely many points of $M_{1}$ (in the same way as $O_{n_{0}}$ from the second paragraph of the proof). Hence, $(M_{1},d_{1})$ is compact, and totally boundedness follows in exactly the same way. For completeness, let $(w_{n})_{n\in{\mathbb{N}}}$ be a Cauchy sequence in $M_{1}$, i.e. we have

As above, $(w_{n})_{n\in{\mathbb{N}}}$ is either eventually constant or provides a sequence $(x_{n})_{n\in{\mathbb{N}}}$ such that $x_{n}\in X_{n}$ for all $n\in{\mathbb{N}}$. The latter case is impossible by assumption and the former case is trivial.

To establish the items in the theorem in $\textup{{ACA}}_{0}^{\omega}+\textup{{QF-AC}}^{0,1}$, the usual proof-by-contradiction goes through as in the proof of Theorem 2.5. ∎

The usual proof-by-contradiction using $\textup{{QF-AC}}^{0,1}$ (and $(\exists^{2})$) shows that sequential uniform continuity implies uniform continuity. For the remaining implications, consider $g:M_{1}\rightarrow{\mathbb{R}}$ from the the proof of Theorem 2.6. This function is sequentially continuous at $0_{M_{1}}$ since any sequence converging to $0_{M_{1}}$ must be eventually constant $0_{M_{1}}$. Similarly, for sequences $(w_{n})_{n\in{\mathbb{N}}}$, $(v_{n})_{n\in{\mathbb{N}}}$ in $M_{1}$ $\lim_{n\rightarrow\infty}d_{1}(w_{n},v_{n})=0$ implies that the sequences are eventually equal. Hence, $g$ is also sequentially uniformly continuous. A similar proof goes through for $(M_{0},d_{0})$ and $f$ from Theorem 2.5. ∎