proof of injective images of Baire space

We show that, for an uncountable Polish spaceMathworldPlanetmath X, there exists a continuousPlanetmathPlanetmath and one-to-one function f:𝒩→X such that Xβˆ–f⁒(𝒩) is countableMathworldPlanetmath. Furthermore, the inversePlanetmathPlanetmathPlanetmathPlanetmath 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 diameterPlanetmathPlanetmath of any subset is defined.


Let S be an uncountable subset of X which can be written as the differencePlanetmathPlanetmath of two closed setsPlanetmathPlanetmath, and choose any Ο΅>0.

Then, there exists a sequenceMathworldPlanetmath S1,S2,… of pairwise disjoint and uncountable subsets of S with diameter no more than Ο΅ and such that,

  1. 1.

    for each n>0, ⋃k≀nSk is closed.

  2. 2.

    Sβˆ–β‹ƒnSn is countable.


As S=Aβˆ–B is the difference of closed sets, it is a Polish space under the subspace topology. In fact, if B is nonempty, the topologyMathworldPlanetmathPlanetmath is generated by the complete metric


otherwise we may take dS=d. In either case, dSβ‰₯d, so it is enough to choose the sets Sn to have diameter no more than 2-n with respect to dS. Note also that any boundedPlanetmathPlanetmathPlanetmathPlanetmath and closed set with respect to this metric is also closed as a subset of X.

Let Sc be the condensation points of S, which are the points whose neighborhoodsMathworldPlanetmathPlanetmath all contain uncountably many points of S. Then, Sβˆ–Sc is a union of countably many countable and open subsets of S, so is countable and open. Hence, Sc is uncountable and closed in S, and every open subset is uncountable. Choosing any p∈Sc then Scβˆ–p will not be a closed subset of X. So, replacing S by Sβˆ–{p} if necessary, we may suppose that Sc is not closed as a subset of X and, therefore is not compactPlanetmathPlanetmath.

So, for some Ξ΄>0, Sc cannot be covered by finitely many sets with dS-diameter no more than Ξ΄ (see here ( By separability, there is a sequence T1,T2,… of open ballsPlanetmathPlanetmath in Sc with diameter less than min⁑(Ο΅,Ξ΄), and covering Sc. Writing TΒ―n for the dS-closureMathworldPlanetmathPlanetmath of Tn and eliminating any terms such that TnβŠ†β‹ƒk<nTΒ―k, then Sn≑TΒ―nβˆ–β‹ƒk<nTΒ―k have nonempty interior and hence are uncountable, and


is countable, as required. ∎

Note that Sn=⋃k≀nSkβˆ–β‹ƒk≀n-1Sk is also a difference of closed sets. So, the lemma allows us to inductively choose sets C⁒(n1,…,nk)βŠ†X for integers kβ‰₯0 and n1,…,nkβ‰₯1 such that C⁒()=X and the following are satisfied.

  1. 1.

    C⁒(n1,…,nk,m) are uncountable, contained in C⁒(n1,…,nk), and pairwise disjoint as m runs through the positive integers.

  2. 2.

    ⋃j≀mC⁒(n1,…,nk,j) is closed for all mβ‰₯1.

  3. 3.

    C⁒(n1,…,nk)βˆ–β‹ƒmC⁒(n1,…,nk,m) is countable and, for kβ‰₯1, has diameter no more than 2-k.

For any nβˆˆπ’© we may choose a sequence xk∈C⁒(n1,…,nk). Since, for kβ‰₯1, this set has diameter no more than 2-k, then d⁒(xj,xk)≀2-k whenever jβ‰₯kβ‰₯1. So, the sequence is Cauchy ( and hence has a limit x. Furthermore, as xj is contained in the closed set


for j>k, then x must also be contained in it and hence is in C⁒(n1,…,nk). So


contains x and is nonempty. Furthermore, as it has zero diameter, it is a singleton. So f:𝒩→X is uniquely defined by f⁒(n)∈C⁒(n).

Given any m,nβˆˆπ’© such that mj=nj for j≀k, then f⁒(m) and f⁒(n) are both in the set C⁒(m1,…,mk), which has diameter no more than 2-k. So, d⁒(f⁒(m),f⁒(n))≀2-k and f is continuous.

If m and n are distinct elements of 𝒩 and k is the smallest integer such that mkβ‰ nk, then f⁒(m) and f⁒(n) are in the disjoint sets C⁒(m1,…,mk) and C⁒(n1,…,nk) respectively. So f⁒(m)β‰ f⁒(n), and f is one-to-one.

Now let A be the countable set


Also define

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

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


Choosing any x∈C⁒(n1,…,nk)βˆ–A then we can inductively find nj for j>k such that x∈C⁒(n1,…,nj). Then, setting n=(n1,n2,…) gives f⁒(n)=x. This shows that

f⁒(𝒩⁒(n1,…,nk))=C⁒(n1,…,nk)βˆ–A. (1)

In particular, f⁒(𝒩)=Xβˆ–A and, therefore Xβˆ–f⁒(𝒩) is countable. Finally, as C⁒(n1,…,nk) 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