Subspaces of Polish Spaces

Use

For a set $A \subseteq X$ and a metric $d$ on $X$, we define the distance of $x$ from $A$ as

Let $U \subseteq X$ be open. Then $d(x , X \setminus U)$ is strictly positive.

By Lemma 1, we can choose a compatible metric $d$ on $X$ with $d < 1$. Let

This is a metric on $U$ compatible with the subspace topology and with respect to which $U$ is complete (exercise!).

Consider the mapping $f: X \to X^\Nat$ given by $x \mapsto (x, x, x, \dots)$. The restriction of $f$ to $\bigcap_n Y_n$ is a homeomorphism between $\bigcap_n Y_n$ and the diagonal $\Delta \subseteq \prod_n Y_n$,

${}\Delta$ is closed in the product space $\prod_n Y_n$ and hence Polish, and this property extends to $\bigcap_n Y_n$ (see Proposition 1).

We can adapt the $\eps$-oscillation set $C_\eps$ used in the proof of Theorem 2 to the domain $A$:

As before, $C^A_\eps$ is open and hence

is $G_\delta$ and since $f$ is continuous, $A \subseteq G \subseteq \Cl{A}$.

To extend $f$ to $G$, let $x \in G$. Since $x \in \Cl{A}$, there exists a sequence $(a_n)$ in $A$ with $x = \lim_n a_n$. As $x \in \bigcap_n C^A_{1/n}$, the sequence $(f(a_n))$ is Cauchy. $Y$ is complete, so there exists $y = \lim_n f(a_n) \in Y$. It is straightforward to verify that $y$ is independent of the choice of $(a_n)$ and agrees with $f(x)$ for $x \in A$. Hence we can put

which yields the desired continuous extension.

By Proposition 1 of the section on Polish spaces and Proposition 1 above, respectively, $F$ and $X \setminus F$ are Polish spaces with compatible metrics $d_F$ and $d_{X\setminus F}$, respectively. By Lemma 1, we may assume $d_F, d_{X\setminus F} < 1$. We form the disjoint union of the spaces $F$ and $X \setminus F$: This is the set $X = F \,\sqcup\, (X \setminus F)$ with the following topology $\mathcal{O}'$: $U \subseteq F \,\sqcup\, (X \setminus F)$ is in $\mathcal{O}'$ if and only if $U \cap F$ is open (in $F$) and $U \cap (X\setminus F)$ is open (in $X\setminus F$).

The disjoint union is Polish, as witnessed by the following metric.

It is straightforward to check that $d_\sqcup$ is compatible with $\mathcal{O}'$. Furthermore, let $(x_n)$ be Cauchy in $(X,d_\sqcup)$. Then the $x_n$ are completely in $F$ or in $X\setminus F$ from some point on, and hence $(x_n)$ converges.

Under the disjoint union topology, $F$ is is clopen. Moreover, an open set in this topology is a disjoint union of an open set in $X\setminus F$, which is also open in the original topology $\mathcal{O}$, and an intersection of an open set from $\mathcal{O}$ with $F$. Such sets are Borel in $(X,\mathcal{O})$ and hence $(X,\mathcal{O})$ and $(X,\mathcal{O}')$ have the same Borel sets.

Let $\mathcal{S}$ be the family of all subsets $A$ of $X$ for which a finer Polish topology exists that has the same Borel sets as $\mathcal{O}$ and in which $A$ is clopen.

We will show that $\mathcal{S}$ is a $\sigma$-algebra, which by the previous lemma contains the closed sets. Hence $\mathcal{S}$ must contain all Borel sets, and we are done.

$\mathcal{S}$ is clearly closed under complements, since the complement of a clopen set is clopen in any topology.

So assume now that $\{A_n\}$ is a countable family of sets in $\mathcal{S}$. Let $\mathcal{O}_n$ be a Polish topology on $X$ that makes $A_n$ clopen and does not introduce new Borel sets.

Let $\mathcal{O}_\infty$ be the topology generated by $\bigcup_n \mathcal{O}_n$. Then $\bigcup_n A_n$ is open in $(X, \mathcal{O}_\infty)$, and we can apply Lemma 3. For this to work, however, we have to show that $(X, \mathcal{O}_\infty)$ is Polish and does not introduce any new Borel sets.

We know that the product space $\prod_n (X,\mathcal{O}_n)$ is Polish. Once again we consider the mapping $\phi: X \to \prod_n X$

Observe that ${}\phi$ is a continuous mapping between $(X,\mathcal{O}_{\infty})$ and $\prod_n X$. The preimage of a basic open set $U_1 \times U_2 \times \cdots \times U_n \times X \times X \times \cdots$ under ${}\phi$ is just the intersection of the $U_i$. Furthermore, ${}\phi$ is clearly one-to-one, and the inverse mapping between $\phi(X)$ and $X$ is continuous, too.

If we can show that $\phi(X)$ is closed in $\prod_n X$, we know it is Polish as a closed subset of a Polish space, and since $(X,\mathcal{O}_\infty)$ is homeomorphic to $\phi(X)$ (in the subspace topology for $\prod_n (X,\mathcal{O}_n)$), we can conclude it is Polish.

To see that $\phi(X)$ is closed in $\prod_n X$, let $(y_1,y_2,y_3, \dots) \in \Co{\phi(X)}$. Then there exist $i < j$ such that $y_i \neq y_j$. Since $(X, \mathcal{O})$ is Polish, we can pick $U,V$ open, disjoint such that $y_i \in U$, $y_j \in V$. Since each $\mathcal{O}_n$ refines $\mathcal{O}$, $U$ is open in $\mathcal{O}_i$, and $V$ is open in $\mathcal{O}_j$. Therefore,

where $X_k = X$ for $k \neq i,j$, is an open neighborhood of $(y_1,y_2,y_3, \dots)$ completely contained in $\Co{\phi(X)}$.

Finally, to see that the Borel sets of $(X, \mathcal{O}_\infty)$ are the same as the ones of $(X,\mathcal{O})$, for each $n$, let $\{U^{(n)}_i\}_{i \in \Nat}$ be a basis for $\mathcal{O}_n$. By assumption, all sets in $\mathcal{O}_n$ are Borel sets of $(X, \mathcal{O})$. The collection $\{U^{(n)}_i\}_{i,n \in \Nat}$ is a subbasis for $\mathcal{O}_\infty$. This means that any open set in $(X, \mathcal{O}_\infty)$ is a countable union of finite intersections of the $U^{(n)}_i$. Since every $U^{(n)}_i$ is Borel in $(X, \mathcal{O})$, this means that any open set in $\mathcal{O}_\infty$ is Borel in $(X, \mathcal{O})$. Since the Borel sets are closed under complementation and countable unions, this in turn implies that every Borel set of $(X, \mathcal{O}_\infty)$ is already Borel in $(X, \mathcal{O})$.

Let $(X,\mathcal{O})$ be Polish, and assume $B \subseteq X$ is Borel. We can choose a finer topology $\mathcal{O}' \supseteq \mathcal{O}$ so that $B$ becomes clopen, but the Borel sets stay the same. By Theorem 1, $B$ is Polish with respect to the subspace topology $\mathcal{O}'|_B$

Suppose $B$ is uncountable. By Theorem 1 there exists a continuous injection $f$ from $\Cant$ (with respect to the standard topology) into $(B,\mathcal{O}'|_B)$.

Since $\mathcal{O}'$ is finer than $\mathcal{O}$, $f$ is continuous with respect to $\mathcal{O}$, too. Since $\Cant$ is compact, $f(\Cant)$ is closed with respect to $\mathcal{O}$. Finally, $f(\Cant)$ has no isolated points with respect to $\mathcal{O}'$, which then also holds for the coarser topology $\mathcal{O}$.

Therefore, $B$ has a perfect subset.