proof of spaces homeomorphic to Baire space

We show that a topological spaceMathworldPlanetmath X is homeomorphicMathworldPlanetmath to Baire spacePlanetmathPlanetmath, 𝒩, if and only if the following are satisfied.

  1. 1.

    It is a nonempty Polish spaceMathworldPlanetmath.

  2. 2.
  3. 3.

    No nonempty and open subsets are compactPlanetmathPlanetmath.

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:𝒩→X. By property 1 there is a complete metric d on X.

We choose subsets C⁒(n1,…,nk) of X for integers kβ‰₯0 and n1,…,nk satisfying the following.

  1. (i)

    C⁒(n1,…,nk) is a nonempty clopen set with diameter no more than 2-k.

  2. (ii)


  3. (iii)

    For any n1,…,nk then C⁒(n1,…,nk,m) are pairwise disjoint as m ranges over the natural numbersMathworldPlanetmath and,

    ⋃m=1∞C⁒(n1,…,nk,m)=C⁒(n1,…,nk). (1)

This can be done inductively. Suppose that S=C⁒(n1,…,nk) has already been chosen. As it is open, condition 3 says that it is not compact. Therefore, there is a Ξ΄>0 such that S has no finite open cover consisting of sets of diameter no more than Ξ΄ (see here ( However, as Polish spaces are separablePlanetmathPlanetmath, there is a countableMathworldPlanetmath sequencePlanetmathPlanetmath S1,S2,… of open sets with diameter less than Ξ΄ and covering S. As the space is zero dimensional, these can be taken to be clopen. By replacing Sj by Sj∩S we can assume that SjβŠ†S. Then, replacing by Sjβˆ–β‹ƒi<jSi, the sets Sj can be taken to be pairwise disjoint.

By eliminating empty setsMathworldPlanetmath we suppose that Sjβ‰ βˆ… for each j, and since S has no finite open cover consisting of sets of diameter less than Ξ΄, the sequence Sj will still be infiniteMathworldPlanetmath. Defining


satisfies the required properties.

We now define a function f:𝒩→X such that f⁒(n)∈C⁒(n1,…,nk) for each nβˆˆπ’© and kβ‰₯0. Choose any nβˆˆπ’© there is a sequence xk∈C⁒(n1,…,nk). This set has diameter boundedPlanetmathPlanetmathPlanetmath by 2-k and, so, d⁒(xk,xj)≀2-k for jβ‰₯k. This sequence is Cauchy ( and, by completeness of the metric, must convergePlanetmathPlanetmath to a limit x. As C⁒(n1,…,nk) is closed, it contains x for each k and therefore

β‹‚kC⁒(n1,…,nk)β‰ βˆ….

In fact, as it has zero diameter, this set must contain a single element, which we define to be f⁒(n).

So, we have defined a function f:𝒩→X. If m,nβˆˆπ’© satisfy mj=nj for j≀k then f⁒(m),f⁒(n) are both contained in C⁒(m1,…,mk) and d⁒(f⁒(m),f⁒(n))≀2-k. Therefore, f is continuousPlanetmathPlanetmath.

It only remains to show that f has continuous inversePlanetmathPlanetmathPlanetmath. Given any x∈X then x∈C⁒() and equation (1) allows us to choose a sequence nkβˆˆβ„• such that x∈C⁒(n1,…,nk) for each k. Then, f⁒(n)=x showing that f is onto.

If mβ‰ nβˆˆπ’© then, letting k be the first integer for which mkβ‰ nk, the sets C⁒(m1,…,mk) and C⁒(n1,…,nk) are disjoint and, therefore, f⁒(m)β‰ f⁒(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

𝒩⁒(n1,…,nk)={mβˆˆπ’©:mj=nj⁒ for ⁒j≀k}

form a basis for the topology on 𝒩. Then, f⁒(𝒩⁒(n1,…,nk))=C⁒(n1,…,nk) is open and, therefore, f is an open map.

Title proof of spaces homeomorphic to Baire space
Canonical name ProofOfSpacesHomeomorphicToBaireSpace
Date of creation 2013-03-22 18:46:51
Last modified on 2013-03-22 18:46:51
Owner gel (22282)
Last modified by gel (22282)
Numerical id 7
Author gel (22282)
Entry type Proof
Classification msc 54E50