proof of injective images of Baire space

We show that, for an uncountable Polish space $X$, there exists a continuous and one-to-one function $f:𝒩\to X$ such that $X\setminus f(𝒩)$ is countable. Furthermore, the inverse of $f$ defined on $f(𝒩)$ is Borel measurable.

The construction of the function relies on the following result. We let $d$ be a complete metric on $X$ with respect to which the diameter of any subset is defined.

Lemma.

Let $S$ be an uncountable subset of $X$ which can be written as the difference of two closed sets, and choose any $ϵ\mathrm{>}\mathrm{0}$.

Then, there exists a sequence $S_{\mathrm{1}}\mathrm{,}{S}_{\mathrm{2}}\mathrm{,}\mathrm{\dots }$ of pairwise disjoint and uncountable subsets of $S$ with diameter no more than $ϵ$ and such that,

1. for each $n>0$, ${\bigcup }_{k\le n}{S}_k$ is closed.

2. $S\setminus {\bigcup }_n{S}_n$ is countable.

Proof.

As $S=A\setminus B$ is the difference of closed sets, it is a Polish space under the subspace topology. In fact, if $B$ is nonempty, the topology is generated by the complete metric

$$d_S(x,y)=d(x,y)+sup\{|d{(x,z)}^{-1}-d{(y,z)}^{-1}|:z\in B\},$$

otherwise we may take $d_S=d$. In either case, $d_S\ge d$, so it is enough to choose the sets $S_n$ to have diameter no more than $2^{-n}$ with respect to $d_S$. Note also that any bounded and closed set with respect to this metric is also closed as a subset of $X$.

Let $S^c$ be the condensation points of $S$, which are the points whose neighborhoods all contain uncountably many points of $S$. Then, $S\setminus S^c$ is a union of countably many countable and open subsets of $S$, so is countable and open. Hence, $S^c$ is uncountable and closed in $S$, and every open subset is uncountable. Choosing any $p\in S^c$ then $S^c\setminus p$ will not be a closed subset of $X$. So, replacing $S$ by $S\setminus \{p\}$ if necessary, we may suppose that $S^c$ is not closed as a subset of $X$ and, therefore is not compact.

So, for some $\delta >0$, $S^c$ cannot be covered by finitely many sets with $d_S$-diameter no more than $\delta$ (see here (http://planetmath.org/ProofThatAMetricSpaceIsCompactIfAndOnlyIfItIsCompleteAndTotallyBounded)). By separability, there is a sequence $T_1,T_2,\mathrm{\dots }$ of open balls in $S^c$ with diameter less than $\min (ϵ,\delta )$, and covering $S^c$. Writing ${\overline{T}}_n$ for the $d_S$-closure of $T_n$ and eliminating any terms such that , then have nonempty interior and hence are uncountable, and

$$S\setminus \bigcup _n{S}_n=S\setminus S^c$$

is countable, as required. ∎

Note that $S_n={\bigcup }_{k\le n}{S}_k\setminus {\bigcup }_{k\le n-1}{S}_k$ is also a difference of closed sets. So, the lemma allows us to inductively choose sets $C(n_1,\mathrm{\dots },n_k)\subseteq X$ for integers $k\ge 0$ and $n_1,\mathrm{\dots },n_k\ge 1$ such that $C()=X$ and the following are satisfied.

1. $C(n_1,\mathrm{\dots },n_k,m)$ are uncountable, contained in $C(n_1,\mathrm{\dots },n_k)$, and pairwise disjoint as $m$ runs through the positive integers.

2. ${\bigcup }_{j\le m}C(n_1,\mathrm{\dots },n_k,j)$ is closed for all $m\ge 1$.

3. $C(n_1,\mathrm{\dots },n_k)\setminus {\bigcup }_{m}C(n_1,\mathrm{\dots },n_k,m)$ is countable and, for $k\ge 1$, has diameter no more than $2^{-k}$.

For any $n\in 𝒩$ we may choose a sequence $x_k\in C(n_1,\mathrm{\dots },n_k)$. Since, for $k\ge 1$, this set has diameter no more than $2^{-k}$, then $d(x_j,x_k)\le 2^{-k}$ whenever $j\ge k\ge 1$. So, the sequence is Cauchy (http://planetmath.org/CauchySequence) and hence has a limit $x$. Furthermore, as $x_j$ is contained in the closed set

$$\bigcup _{m\le n_{k+1}}C(n_1,\mathrm{\dots },n_k,m)$$

for $j>k$, then $x$ must also be contained in it and hence is in $C(n_1,\mathrm{\dots },n_k)$. So

$$C(n)\equiv \bigcap _{k=0}^{\mathrm{\infty }}C(n_1,\mathrm{\dots },n_k)$$

contains $x$ and is nonempty. Furthermore, as it has zero diameter, it is a singleton. So $f:𝒩\to X$ is uniquely defined by $f(n)\in C(n)$.

Given any $m,n\in 𝒩$ such that $m_j=n_j$ for $j\le k$, then $f(m)$ and $f(n)$ are both in the set $C(m_1,\mathrm{\dots },m_k)$, which has diameter no more than $2^{-k}$. So, $d(f(m),f(n))\le 2^{-k}$ and $f$ is continuous.

If $m$ and $n$ are distinct elements of $𝒩$ and $k$ is the smallest integer such that $m_k\ne n_k$, then $f(m)$ and $f(n)$ are in the disjoint sets $C(m_1,\mathrm{\dots },m_k)$ and $C(n_1,\mathrm{\dots },n_k)$ respectively. So $f(m)\ne f(n)$, and $f$ is one-to-one.

Now let $A$ be the countable set

$$A=\bigcup _k\bigcup _{n_1,\mathrm{\dots },n_k}\left(C(n_1,\mathrm{\dots },n_k)\setminus \bigcup _{m}C(n_1,\mathrm{\dots },n_k,m)\right)\subseteq X.$$

Also define

$$𝒩(n_1,\mathrm{\dots },n_k)\equiv \{m\in 𝒩:m_j=n_j\text{ for }j\le k\}.$$

These sets form a basis for the topology on $𝒩$. Clearly,

$$f\left(𝒩(n_1,\mathrm{\dots },n_k)\right)\subseteq C(n_1,\mathrm{\dots },n_k)\setminus A.$$

Choosing any $x\in C(n_1,\mathrm{\dots },n_k)\setminus A$ then we can inductively find $n_j$ for $j>k$ such that $x\in C(n_1,\mathrm{\dots },n_j)$. Then, setting $n=(n_1,n_2,\mathrm{\dots })$ gives $f(n)=x$. This shows that

$$f\left(𝒩(n_1,\mathrm{\dots },n_k)\right)=C(n_1,\mathrm{\dots },n_k)\setminus A.$$ (1)

In particular, $f(𝒩)=X\setminus A$ and, therefore $X\setminus f(𝒩)$ is countable. Finally, as $C(n_1,\mathrm{\dots },n_k)$ is a difference of closed sets, it is Borel, and equation (1) shows that the inverse of $f$ is Borel measurable.

Title	proof of injective images of Baire space
Canonical name	ProofOfInjectiveImagesOfBaireSpace
Date of creation	2013-03-22 18:47:15
Last modified on	2013-03-22 18:47:15
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	5
Author	gel (22282)
Entry type	Proof
Classification	msc 54E50