# 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 $f\colon\mathcal{N}\rightarrow X$ such that $X\setminus f(\mathcal{N})$ is countable. Furthermore, the inverse of $f$ defined on $f(\mathcal{N})$ 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 $\epsilon>0$.

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

1. 1.

for each $n>0$, $\bigcup_{k\leq n}S_{k}$ is closed.

2. 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

 $d_{S}(x,y)=d(x,y)+\sup\left\{|d(x,z)^{-1}-d(y,z)^{-1}|\colon z\in B\right\},$

otherwise we may take $d_{S}=d$. In either case, $d_{S}\geq d$, 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 $T_{1},T_{2},\ldots$ of open balls in $S^{c}$ with diameter less than $\min(\epsilon,\delta)$, and covering $S^{c}$. Writing $\bar{T}_{n}$ for the $d_{S}$-closure of $T_{n}$ and eliminating any terms such that $T_{n}\subseteq\bigcup_{k, then $S_{n}\equiv\bar{T}_{n}\setminus\bigcup_{k have nonempty interior and hence are uncountable, and

 $S\setminus\bigcup_{n}S_{n}=S\setminus S^{c}$

is countable, as required. ∎

Note that $S_{n}=\bigcup_{k\leq n}S_{k}\setminus\bigcup_{k\leq n-1}S_{k}$ is also a difference of closed sets. So, the lemma allows us to inductively choose sets $C(n_{1},\ldots,n_{k})\subseteq X$ for integers $k\geq 0$ and $n_{1},\ldots,n_{k}\geq 1$ such that $C()=X$ and the following are satisfied.

1. 1.

$C(n_{1},\ldots,n_{k},m)$ are uncountable, contained in $C(n_{1},\ldots,n_{k})$, and pairwise disjoint as $m$ runs through the positive integers.

2. 2.

$\bigcup_{j\leq m}C(n_{1},\ldots,n_{k},j)$ is closed for all $m\geq 1$.

3. 3.

$C(n_{1},\ldots,n_{k})\setminus\bigcup_{m}C(n_{1},\ldots,n_{k},m)$ is countable and, for $k\geq 1$, has diameter no more than $2^{-k}$.

For any $n\in\mathcal{N}$ we may choose a sequence $x_{k}\in C(n_{1},\ldots,n_{k})$. Since, for $k\geq 1$, this set has diameter no more than $2^{-k}$, then $d(x_{j},x_{k})\leq 2^{-k}$ whenever $j\geq k\geq 1$. 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

 $\bigcup_{m\leq n_{k+1}}C(n_{1},\ldots,n_{k},m)$

for $j>k$, then $x$ must also be contained in it and hence is in $C(n_{1},\ldots,n_{k})$. So

 $C(n)\equiv\bigcap_{k=0}^{\infty}C(n_{1},\ldots,n_{k})$

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

Given any $m,n\in\mathcal{N}$ such that $m_{j}=n_{j}$ for $j\leq k$, then $f(m)$ and $f(n)$ are both in the set $C(m_{1},\ldots,m_{k})$, which has diameter no more than $2^{-k}$. So, $d(f(m),f(n))\leq 2^{-k}$ and $f$ is continuous.

If $m$ and $n$ are distinct elements of $\mathcal{N}$ and $k$ is the smallest integer such that $m_{k}\not=n_{k}$, then $f(m)$ and $f(n)$ are in the disjoint sets $C(m_{1},\ldots,m_{k})$ and $C(n_{1},\ldots,n_{k})$ respectively. So $f(m)\not=f(n)$, and $f$ is one-to-one.

Now let $A$ be the countable set

 $A=\bigcup_{k}\bigcup_{n_{1},\ldots,n_{k}}\left(C(n_{1},\ldots,n_{k})\setminus% \bigcup_{m}C(n_{1},\ldots,n_{k},m)\right)\subseteq X.$

Also define

 $\mathcal{N}(n_{1},\ldots,n_{k})\equiv\left\{m\in\mathcal{N}\colon m_{j}=n_{j}% \text{ for }j\leq k\right\}.$

These sets form a basis for the topology on $\mathcal{N}$. Clearly,

 $f\left(\mathcal{N}(n_{1},\ldots,n_{k})\right)\subseteq C(n_{1},\ldots,n_{k})% \setminus A.$

Choosing any $x\in C(n_{1},\ldots,n_{k})\setminus A$ then we can inductively find $n_{j}$ for $j>k$ such that $x\in C(n_{1},\ldots,n_{j})$. Then, setting $n=(n_{1},n_{2},\ldots)$ gives $f(n)=x$. This shows that

 $f\left(\mathcal{N}(n_{1},\ldots,n_{k})\right)=C(n_{1},\ldots,n_{k})\setminus A.$ (1)

In particular, $f(\mathcal{N})=X\setminus A$ and, therefore $X\setminus f(\mathcal{N})$ is countable. Finally, as $C(n_{1},\ldots,n_{k})$ 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 ProofOfInjectiveImagesOfBaireSpace 2013-03-22 18:47:15 2013-03-22 18:47:15 gel (22282) gel (22282) 5 gel (22282) Proof msc 54E50