proof of spaces homeomorphic to Baire space
We show that a topological space^{} $X$ is homeomorphic^{} to Baire space^{}, $\mathrm{\pi \x9d\x92\copyright}$, if and only if the following are satisfied.

1.
It is a nonempty Polish space^{}.

2.
It is zero dimensional.

3.
No nonempty and open subsets are compact^{}.
As Baire space is easily shown to satisfy these properties, we just need to show that if they are satisfied then there exists a homeomorphism $f:\mathrm{\pi \x9d\x92\copyright}\beta \x86\x92X$. By property 1 there is a complete metric $d$ on $X$.
We choose subsets $C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$ of $X$ for integers $k\beta \x89\u20af0$ and ${n}_{1},\mathrm{\beta \x80\xa6},{n}_{k}$ satisfying the following.

(i)
$C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$ is a nonempty clopen set with diameter no more than ${2}^{k}$.

(ii)
$C\beta \x81\u2019()=X$.

(iii)
For any ${n}_{1},\mathrm{\beta \x80\xa6},{n}_{k}$ then $C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k},m)$ are pairwise disjoint as $m$ ranges over the natural numbers^{} and,
$$\underset{m=1}{\overset{\mathrm{\beta \x88\x9e}}{\beta \x8b\x83}}C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k},m)=C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k}).$$ (1)
This can be done inductively. Suppose that $S=C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$ has already been chosen. As it is open, condition 3 says that it is not compact. Therefore, there is a $\mathrm{\Xi \u0384}>0$ such that $S$ has no finite open cover consisting of sets of diameter no more than $\mathrm{\Xi \u0384}$ (see here (http://planetmath.org/ProofThatAMetricSpaceIsCompactIfAndOnlyIfItIsCompleteAndTotallyBounded)). However, as Polish spaces are separable^{}, there is a countable^{} sequence^{} ${S}_{1},{S}_{2},\mathrm{\beta \x80\xa6}$ of open sets with diameter less than $\mathrm{\Xi \u0384}$ and covering $S$. As the space is zero dimensional, these can be taken to be clopen. By replacing ${S}_{j}$ by ${S}_{j}\beta \x88\copyright S$ we can assume that ${S}_{j}\beta \x8a\x86S$. Then, replacing by $$, the sets ${S}_{j}$ can be taken to be pairwise disjoint.
By eliminating empty sets^{} we suppose that ${S}_{j}\beta \x89\mathrm{\beta \x88\x85}$ for each $j$, and since $S$ has no finite open cover consisting of sets of diameter less than $\mathrm{\Xi \u0384}$, the sequence ${S}_{j}$ will still be infinite^{}. Defining
$$C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k},{n}_{k+1})={S}_{{n}_{k+1}}$$ 
satisfies the required properties.
We now define a function $f:\mathrm{\pi \x9d\x92\copyright}\beta \x86\x92X$ such that $f\beta \x81\u2019(n)\beta \x88\x88C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$ for each $n\beta \x88\x88\mathrm{\pi \x9d\x92\copyright}$ and $k\beta \x89\u20af0$. Choose any $n\beta \x88\x88\mathrm{\pi \x9d\x92\copyright}$ there is a sequence ${x}_{k}\beta \x88\x88C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$. This set has diameter bounded^{} by ${2}^{k}$ and, so, $d\beta \x81\u2019({x}_{k},{x}_{j})\beta \x89\u20ac{2}^{k}$ for $j\beta \x89\u20afk$. This sequence is Cauchy (http://planetmath.org/CauchySequence) and, by completeness of the metric, must converge^{} to a limit $x$. As $C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$ is closed, it contains $x$ for each $k$ and therefore
$$\underset{k}{\beta \x8b\x82}C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\beta \x89\mathrm{\beta \x88\x85}.$$ 
In fact, as it has zero diameter, this set must contain a single element, which we define to be $f\beta \x81\u2019(n)$.
So, we have defined a function $f:\mathrm{\pi \x9d\x92\copyright}\beta \x86\x92X$. If $m,n\beta \x88\x88\mathrm{\pi \x9d\x92\copyright}$ satisfy ${m}_{j}={n}_{j}$ for $j\beta \x89\u20ack$ then $f\beta \x81\u2019(m),f\beta \x81\u2019(n)$ are both contained in $C\beta \x81\u2019({m}_{1},\mathrm{\beta \x80\xa6},{m}_{k})$ and $d\beta \x81\u2019(f\beta \x81\u2019(m),f\beta \x81\u2019(n))\beta \x89\u20ac{2}^{k}$. Therefore, $f$ is continuous^{}.
It only remains to show that $f$ has continuous inverse^{}. Given any $x\beta \x88\x88X$ then $x\beta \x88\x88C\beta \x81\u2019()$ and equation (1) allows us to choose a sequence ${n}_{k}\beta \x88\x88\mathrm{\beta \x84\x95}$ such that $x\beta \x88\x88C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$ for each $k$. Then, $f\beta \x81\u2019(n)=x$ showing that $f$ is onto.
If $m\beta \x89n\beta \x88\x88\mathrm{\pi \x9d\x92\copyright}$ then, letting $k$ be the first integer for which ${m}_{k}\beta \x89{n}_{k}$, the sets $C\beta \x81\u2019({m}_{1},\mathrm{\beta \x80\xa6},{m}_{k})$ and $C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$ are disjoint and, therefore, $f\beta \x81\u2019(m)\beta \x89f\beta \x81\u2019(n)$ and $f$ is one to one.
Finally, we show that $f$ is an open map, so that its inverse is continuous. Sets of the form
$$\mathrm{\pi \x9d\x92\copyright}\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})=\{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\}$$ 
form a basis for the topology on $\mathrm{\pi \x9d\x92\copyright}$. Then, $f\beta \x81\u2019\left(\mathrm{\pi \x9d\x92\copyright}\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})\right)=C\beta \x81\u2019({n}_{1},\mathrm{\beta \x80\xa6},{n}_{k})$ is open and, therefore, $f$ is an open map.
Title  proof of spaces homeomorphic to Baire space 

Canonical name  ProofOfSpacesHomeomorphicToBaireSpace 
Date of creation  20130322 18:46:51 
Last modified on  20130322 18:46:51 
Owner  gel (22282) 
Last modified by  gel (22282) 
Numerical id  7 
Author  gel (22282) 
Entry type  Proof 
Classification  msc 54E50 