5

My class notes say that because $U=\mathbb{R}\backslash\{x_1,x_2,..,x_n\}$ has the same cardinality than $\mathbb{R}$, there exists a homeomorphism between:

$(U,T_{cof})$ and $(\mathbb{R},T_{cof})$, where $T_{cof}$ is the finite complement topology.

I initially thought that having the same cardinality is a necessary condition but is not sufficient to have an homeomorphism. Also, I can't manage to find a homeomorphism between these two.

Benjamin Dickman
  • 14,451
Pao
  • 225

3 Answers3

2

You are right that in general having the same cardinality is not enough (e.g. $[0,1]$ vs $(0,1)$ with Euclidean topology). However in the case of finite complement topology it is enough.

For that consider any bijection $f:X\to Y$ with finite complement topology on both sides. It is continuous because the preimage of a finite set is finite. It is closed because the image of a finite set is finite. These two properties are enough to ensure that $f$ is a homeomorphism.

freakish
  • 42,851
  • Ok, thank you very much for your answer. However it seems difficult to find a bijection in practice.. (although there must be one because they have the same cardinality!) – Pao Jul 01 '18 at 09:02
  • @Pao Yes, having the same cardinality means (by definition) that such bijection exists. Constructing such bijections may or may not be easy. I didn't put much thought into this. – freakish Jul 01 '18 at 09:06
  • I think for the continuity of $f$, we should say "because the presage of a infinite set is infinite". The reason is that, if we let $A$ be an infinite set in $Y$, then $[f^{-1}(A)]^c$ may not equal $f^{-1}(A^c)$. So we can't conclude the continuity of $f$ from "the preimage of a finite set is finite". – Sam Wong Oct 26 '18 at 06:02
  • @SamWong we can because a subset is closed if and only if it is finite. I'm using "preimage of closed sets" continuity. Also preimage of complement is complement of preimage under bijection. – freakish Oct 26 '18 at 06:04
  • @freakish Oh yeah, I didn't see the bijection condition. You are right. I am thinking of the open sets version proposition. – Sam Wong Oct 26 '18 at 06:09
2

Enumerate a countably infinite subset $\{a_k\}_{k \in \mathbb{N}}$ of $\mathbb{R}$ for which the first $n$ terms are $x_1$ through $x_n$. Here is an explicit homeomorphism from $\mathbb{R} \rightarrow U$ where the latter is endowed with the finite complement topology:

If $x = x_k$, then map $x \mapsto a_{k+n}$; otherwise, use the identity mapping $x \mapsto x$.

I will leave it to you to check that this function is one-to-one [injective] and onto [surjective].

To show that it is continuous, it suffices to show that the inverse image of any closed set is closed. The closed sets in $U$ are exactly finite subsets of $U$; can you show that the inverse image of such a closed subset is also finite in $\mathbb{R}$ hence closed in $\mathbb{R}$ under the finite complement topology?

The last step is to show that the inverse of this function is continuous, too; if you can reason out the above, then the last step should be doable, too.

Benjamin Dickman
  • 14,451
  • Thank you! I agree with your answer, except for the step when you mention that closed sets in V are exactly finite subsets of {1,…,N}. In my opinion, closed sets in V must rather contain {1,...,N}. – Pao Jul 01 '18 at 09:48
  • @Pao Corrected the typo! The closed sets in $V$ are precisely its finite subsets. – Benjamin Dickman Jul 01 '18 at 17:06
  • You have some errors. In your first sentence you talk about a mapping that "does not compute" since, say, $x_1 \notin U$. – CopyPasteIt Jul 01 '18 at 17:57
  • @CopyPasteIt Thanks! I've edited to avoid that issue (and make the proof a bit more direct). – Benjamin Dickman Jul 01 '18 at 18:19
  • 1
    Of course on a quick read, we just nod our heads and say, "sure - that looks reasonable!" (for prior version). $\quad \text{ :) }$ – CopyPasteIt Jul 01 '18 at 19:36
1

The following is easy to prove:

Proposition 1: Let $X$ be a set endowed with the cofinite topology. The subspace topology on any $Y \subset X$ is identical to the cofinite topology on $Y$.
Proof: Exercise.

Proposition 2: Let $X$ and $Y$ be two cofinite topological spaces. Then any injective mapping $f: X \to Y$ is continuous.
Proof
The proof will be complete if we can show that

$\tag 1 \text{For any } A \subset X, \; f(\overline{A})\subseteq \overline{f(A)}$

(click here).

There are two cases:

Case 1: If $A$ is infinite, $\overline{A} = X$. Since $f$ is injective, $f(A)$ is infinite and $\overline{f(A)} = Y$.
Since $f(X) \subseteq Y$, $\text{(1)}$ must be true.

Case 2: If $A$ is finite, $\overline{A} = A$ and the lhs of $\text{(1)}$ is $f(A)$. The image $f(A)$ is also finite and the rhs of $\text{(1)}$ is also $f(A)$, and so the inclusion relation in $\text{(1)}$ must again be true.$\quad \blacksquare$

Using the above we can now state

Proposition 3: Let $f$ be an injective mapping from a set $X$ with the cofinite topology to a set $Y$ with the cofinite topology. Then $f$ is a homeomorphism between $X$ and its image $f(X)$.

Let $U=\mathbb{R}\backslash\{x_1,x_2,..,x_n\}$ for distinct numbers $x_i$ and assume that

$\quad x_n = \text{max(}x_1,x_2,..,x_n\text{)}$

Extend the finite sequence by defining $x_{n+k} = x_n + k$ for $k \ge 1$.

We can define an injection on the set of $x_i$ via

$\quad x_i \to x_{i+n}$

This injection can be easily extended to define a bijective correspondence between $\mathbb R$ and $U$.

CopyPasteIt
  • 11,366
  • 1
    As a piece of trivia: The last idea, for the countably infinite case, dates back at least to the correspondence between Dedekind and Cantor. See Gouvea's AMM article (cf. from the second full paragraph on p. 204: bijecting the irrational numbers in the unit interval with the entire unit interval). The finite case of bijecting $(0, 1]$ and $[0, 1]$ is mentioned there, too. [Anyway: +1] – Benjamin Dickman Jul 01 '18 at 18:31