proof of injective images of Baire space
We show that, for an uncountable Polish space^{} $X$, there exists a continuous^{} and onetoone function $f:\mathrm{\pi \x9d\x92\copyright}\beta \x86\x92X$ such that $X\beta \x88\x96f\beta \x81\u2019(\mathrm{\pi \x9d\x92\copyright})$ is countable^{}. Furthermore, the inverse^{} of $f$ defined on $f\beta \x81\u2019(\mathrm{\pi \x9d\x92\copyright})$ 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 $\mathrm{{\rm O}\u0385}\mathrm{>}\mathrm{0}$.
Then, there exists a sequence^{} ${S}_{\mathrm{1}}\mathrm{,}{S}_{\mathrm{2}}\mathrm{,}\mathrm{\beta \x80\xa6}$ of pairwise disjoint and uncountable subsets of $S$ with diameter no more than $\mathrm{{\rm O}\u0385}$ and such that,

1.
for each $n>0$, ${\beta \x8b\x83}_{k\beta \x89\u20acn}{S}_{k}$ is closed.

2.
$S\beta \x88\x96{\beta \x8b\x83}_{n}{S}_{n}$ is countable.
Proof.
As $S=A\beta \x88\x96B$ 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
$${d}_{S}\beta \x81\u2019(x,y)=d\beta \x81\u2019(x,y)+sup\beta \x81\u2018\{d\beta \x81\u2019{(x,z)}^{1}d\beta \x81\u2019{(y,z)}^{1}:z\beta \x88\x88B\},$$ 
otherwise we may take ${d}_{S}=d$. In either case, ${d}_{S}\beta \x89\u20afd$, 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, $S\beta \x88\x96{S}^{c}$ 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 $p\beta \x88\x88{S}^{c}$ then ${S}^{c}\beta \x88\x96p$ will not be a closed subset of $X$. So, replacing $S$ by $S\beta \x88\x96\{p\}$ if necessary, we may suppose that ${S}^{c}$ is not closed as a subset of $X$ and, therefore is not compact^{}.
So, for some $\mathrm{\Xi \u0384}>0$, ${S}^{c}$ cannot be covered by finitely many sets with ${d}_{S}$diameter no more than $\mathrm{\Xi \u0384}$ (see here (http://planetmath.org/ProofThatAMetricSpaceIsCompactIfAndOnlyIfItIsCompleteAndTotallyBounded)). By separability, there is a sequence ${T}_{1},{T}_{2},\mathrm{\beta \x80\xa6}$ of open balls^{} in ${S}^{c}$ with diameter less than $\mathrm{min}\beta \x81\u2018(\mathrm{{\rm O}\u0385},\mathrm{\Xi \u0384})$, and covering ${S}^{c}$. Writing ${\stackrel{{\rm B}\u2015}{T}}_{n}$ for the ${d}_{S}$closure^{} of ${T}_{n}$ and eliminating any terms such that $$, then $$ have nonempty interior and hence are uncountable, and
$$S\beta \x88\x96\underset{n}{\beta \x8b\x83}{S}_{n}=S\beta \x88\x96{S}^{c}$$ 
is countable, as required. β
Note that ${S}_{n}={\beta \x8b\x83}_{k\beta \x89\u20acn}{S}_{k}\beta \x88\x96{\beta \x8b\x83}_{k\beta \x89\u20acn1}{S}_{k}$ is also a difference of closed sets. So, the lemma allows us to inductively choose sets $C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\beta \x8a\x86X$ for integers $k\beta \x89\u20af0$ and ${n}_{1},\mathrm{\beta \x80\xa6},{n}_{k}\beta \x89\u20af1$ such that $C\beta \x81\u2019()=X$ and the following are satisfied.

1.
$C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k},m)$ are uncountable, contained in $C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$, and pairwise disjoint as $m$ runs through the positive integers.

2.
${\beta \x8b\x83}_{j\beta \x89\u20acm}C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k},j)$ is closed for all $m\beta \x89\u20af1$.

3.
$C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\beta \x88\x96{\beta \x8b\x83}_{m}C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k},m)$ is countable and, for $k\beta \x89\u20af1$, has diameter no more than ${2}^{k}$.
For any $n\beta \x88\x88\mathrm{\pi \x9d\x92\copyright}$ we may choose a sequence ${x}_{k}\beta \x88\x88C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$. Since, for $k\beta \x89\u20af1$, this set has diameter no more than ${2}^{k}$, then $d\beta \x81\u2019({x}_{j},{x}_{k})\beta \x89\u20ac{2}^{k}$ whenever $j\beta \x89\u20afk\beta \x89\u20af1$. 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
$$\underset{m\beta \x89\u20ac{n}_{k+1}}{\beta \x8b\x83}C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k},m)$$ 
for $j>k$, then $x$ must also be contained in it and hence is in $C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$. So
$$C\beta \x81\u2019(n)\beta \x89\u2018\underset{k=0}{\overset{\mathrm{\beta \x88\x9e}}{\beta \x8b\x82}}C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$$ 
contains $x$ and is nonempty. Furthermore, as it has zero diameter, it is a singleton. So $f:\mathrm{\pi \x9d\x92\copyright}\beta \x86\x92X$ is uniquely defined by $f\beta \x81\u2019(n)\beta \x88\x88C\beta \x81\u2019(n)$.
Given any $m,n\beta \x88\x88\mathrm{\pi \x9d\x92\copyright}$ such that ${m}_{j}={n}_{j}$ for $j\beta \x89\u20ack$, then $f\beta \x81\u2019(m)$ and $f\beta \x81\u2019(n)$ are both in the set $C\beta \x81\u2019({m}_{1},\mathrm{\beta \x80\xa6},{m}_{k})$, which has diameter no more than ${2}^{k}$. So, $d\beta \x81\u2019(f\beta \x81\u2019(m),f\beta \x81\u2019(n))\beta \x89\u20ac{2}^{k}$ and $f$ is continuous.
If $m$ and $n$ are distinct elements of $\mathrm{\pi \x9d\x92\copyright}$ and $k$ is the smallest integer such that ${m}_{k}\beta \x89{n}_{k}$, then $f\beta \x81\u2019(m)$ and $f\beta \x81\u2019(n)$ are in the disjoint sets $C\beta \x81\u2019({m}_{1},\mathrm{\beta \x80\xa6},{m}_{k})$ and $C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$ respectively. So $f\beta \x81\u2019(m)\beta \x89f\beta \x81\u2019(n)$, and $f$ is onetoone.
Now let $A$ be the countable set
$$A=\underset{k}{\beta \x8b\x83}\underset{{n}_{1},\mathrm{\beta \x80\xa6},{n}_{k}}{\beta \x8b\x83}\left(C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\beta \x88\x96\underset{m}{\beta \x8b\x83}C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k},m)\right)\beta \x8a\x86X.$$ 
Also define
$$\mathrm{\pi \x9d\x92\copyright}\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\beta \x89\u2018\{m\beta \x88\x88\mathrm{\pi \x9d\x92\copyright}:{m}_{j}={n}_{j}\beta \x81\u2019\text{\Beta for\Beta}\beta \x81\u2019j\beta \x89\u20ack\}.$$ 
These sets form a basis for the topology on $\mathrm{\pi \x9d\x92\copyright}$. Clearly,
$$f\beta \x81\u2019\left(\mathrm{\pi \x9d\x92\copyright}\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\right)\beta \x8a\x86C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\beta \x88\x96A.$$ 
Choosing any $x\beta \x88\x88C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\beta \x88\x96A$ then we can inductively find ${n}_{j}$ for $j>k$ such that $x\beta \x88\x88C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{j})$. Then, setting $n=({n}_{1},{n}_{2},\mathrm{\beta \x80\xa6})$ gives $f\beta \x81\u2019(n)=x$. This shows that
$$f\beta \x81\u2019\left(\mathrm{\pi \x9d\x92\copyright}\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\right)=C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\beta \x88\x96A.$$  (1) 
In particular, $f\beta \x81\u2019(\mathrm{\pi \x9d\x92\copyright})=X\beta \x88\x96A$ and, therefore $X\beta \x88\x96f\beta \x81\u2019(\mathrm{\pi \x9d\x92\copyright})$ is countable. Finally, as $C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$ 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  20130322 18:47:15 
Last modified on  20130322 18:47:15 
Owner  gel (22282) 
Last modified by  gel (22282) 
Numerical id  5 
Author  gel (22282) 
Entry type  Proof 
Classification  msc 54E50 