Return to Answer

Proof. Let $f$ be such a Borel map. In particular, $f(x)=f(y)$ whenever $g \cdot x = y$ for some $g \in Fin(\mathbb{N})$ where the action is given by permuting the indices.

Each $g \in Fin(\mathbb{N})$ induces a homeomorphism of ${\overline{\mathbb{R}}}^{\mathbb{N}}$. Then, since $f$ is producing the same element for sequences that are in the same orbit under the action of $Fin(\mathbb{N})$, $g[A_q] \triangle g[U]=A_q \triangle g[U]$. So, each $A_q \triangle g[U]$ is meager and so is $A_q \triangle \bigcup_{g} g[U]$.

From the proof it can be seen that if we weaken the uniformity condition on $f$ so that it produces the same element for the sequences that are $Fin(\mathbb{N})$-equivalent, we still cannot diagonalize in a Borel way.

Proof. Let $f$ be such a Borel map. In particular, $f(x)=f(y)$ whenever $g \cdot x = y$ for some $g \in \operatorname{Fin}(\mathbb{N})$ where the action is given by permuting the indices.

Each $g \in \operatorname{Fin}(\mathbb{N})$ induces a homeomorphism of ${\overline{\mathbb{R}}}^{\mathbb{N}}$. Then, since $f$ is producing the same element for sequences that are in the same orbit under the action of $\operatorname{Fin}(\mathbb{N})$, $g[A_q] \triangle g[U]=A_q \triangle g[U]$. So, each $A_q \triangle g[U]$ is meager and so is $A_q \triangle \bigcup_{g} g[U]$.

From the proof it can be seen that if we weaken the uniformity condition on $f$ so that it produces the same element for the sequences that are $\operatorname{Fin}(\mathbb{N})$-equivalent, we still cannot diagonalize in a Borel way.

Proof. Pick some $q \in \mathbb{Q}$. Then, $A_q$ is also Borel ${\overline{\mathbb{R}}}^{\mathbb{N}}$ and hence has the Baire property. Let $U$ be an open set such that $A_q \triangle U$ is meager.

Proof. Pick some $q \in \mathbb{Q}$. Then, $A_q$ is also Borel in ${\overline{\mathbb{R}}}^{\mathbb{N}}$ and hence has the Baire property. Let $U$ be an open set such that $A_q \triangle U$ is meager.

Each $g \in Fin(\mathbb{N})$ induces a homeomorphism of ${\overline{\mathbb{R}}}^{\mathbb{N}}$. Then, since $f$ preserves images of sequences that are in the same orbit under the action of $Fin(\mathbb{N})$, each $g[A_q] \triangle g[U]=A_q \triangle g[U]$ is meager and so is $A_q \triangle \bigcup_{g} g[U]$.

If $U=\emptyset$, then $A_q$ is meager. If $U \neq \emptyset$, then $\bigcup_{g} g[U]$ is dense and $A_q^C$ is meager (since $(\bigcup_{g} g[U])^C$ and $\bigcup_{g} g[U] - A_q$ are meager). This completes the proof of the lemma.

Now, let $S$ be the set of all rationals $q$ such that $A_q$ is meager in ${\overline{\mathbb{R}}}^{\mathbb{N}}$. $S$ cannot be $\emptyset$ or $\mathbb{Q}$ by the Baire category theorem.

  Let $z=sup(S)$. Then, since $\bigcup_{s < z} A_q \cup \bigcup_{s > z} A_q^C$ is meager, $\{x \in {\mathbb{R}}^{\mathbb{N}}: f(x) = z\}$ would be comeager, and hence dense the BCT. But then, $\{x \in {\mathbb{R}}^{\mathbb{N}}: f(x) = z\}$ being dense implies that it has an element starting with $z$. $\blacksquare$

From the proof it can be seen that if we weaken the uniformity condition on $f$ so that it only preserves the images under the action of $Fin(\mathbb{N})$, we still cannot diagonalize in a Borel way.

The operation $E \mapsto E^+$ is known as the Friedman-Stanley jump. For any Borel equivalence relation $E$ with more than one equivalence class on a standard Borel space, $E <_B E^+$ where $\leq_B$ denotes Borel reducibility. Indeed, Friedman's own proof that $^+$ is a jump operator makes use of the Borel diagonalization theorem.

As one last remark, I should add that in the same paper, Friedman claims that in order to prove this Borel diaganolization theorem in its full form, it is necessary to use $\omega_1$ many iterations of the power set of operation (Corollary 3.4), just like Borel determinacy. Indeed, according to Lemma 3.2.5, there exists a constant $n \in \omega$ such that Borel determinacy up to Borel sets of rank $ \leq \alpha+n$ implies the Borel diagonalization theorem for Borel equivalence relations of rank $\leq \alpha$.

Each $g \in Fin(\mathbb{N})$ induces a homeomorphism of ${\overline{\mathbb{R}}}^{\mathbb{N}}$. Then, since $f$ is producing the same element for sequences that are in the same orbit under the action of $Fin(\mathbb{N})$, $g[A_q] \triangle g[U]=A_q \triangle g[U]$. So, each $A_q \triangle g[U]$ is meager and so is $A_q \triangle \bigcup_{g} g[U]$.

If $U=\emptyset$, then $A_q$ is meager. If $U \neq \emptyset$, then $\bigcup_{g} g[U]$ is dense and it follows that $A_q^C$ is meager since $(\bigcup_{g} g[U])^C$ and $\bigcup_{g} g[U] - A_q$ are meager. This completes the proof of the lemma. $\blacksquare$

Now, let $S$ be the set of all rationals $q$ such that $A_q$ is meager in ${\overline{\mathbb{R}}}^{\mathbb{N}}$. $S$ cannot be $\emptyset$ or $\mathbb{Q}$ by the Baire category theorem. Let $z=sup(S)$. Then, since $\bigcup_{s < z} A_q \cup \bigcup_{s > z} A_q^C$ is meager, $\{x \in {\mathbb{R}}^{\mathbb{N}}: f(x) = z\}$ is comeager, and hence dense by the BCT. But then, $\{x \in {\mathbb{R}}^{\mathbb{N}}: f(x) = z\}$ being dense implies that it has an element starting with $z$. $\blacksquare$

From the proof it can be seen that if we weaken the uniformity condition on $f$ so that it produces the same element for the sequences that are $Fin(\mathbb{N})$-equivalent, we still cannot diagonalize in a Borel way.

The operation $E \mapsto E^+$ is known as the Friedman-Stanley jump. For any Borel equivalence relation $E$ with more than one equivalence class on a standard Borel space, $E <_B E^+$ where $\leq_B$ denotes the Borel reducibility. Indeed, Friedman's own proof that $^+$ is a jump operator makes use of the Borel diagonalization theorem.

As a last remark, I should add that in the above paper Friedman claims that in order to prove this Borel diaganolization theorem in its full form, it is necessary to use $\omega_1$ many iterations of the power set of operation (Corollary 3.4), just like Borel determinacy theorem. Indeed, according to Lemma 3.2.5, there exists a constant $n \in \omega$ such that Borel determinacy up to Borel sets of rank $ \leq \alpha+n$ implies the Borel diagonalization theorem for Borel equivalence relations of rank $\leq \alpha$.