Topology: Homeomorphism between finite complement topology in $\mathbb{R}$ and one of its subspaces

Question

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.

Answer 1

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.

Answer 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.

Answer 3

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 $U$.