# Cofinite\discrete subspace of a T1 space?

Let $$(X,\tau)$$ be a $$T_1$$-space and $$X$$ is an infinite set. Then $$(X,\tau)$$ has a subspace homeomorphic to $$(\mathbb{N},\tau_2)$$, where $$\tau_2$$ is either the finite-closed topology or the discrete topology.

Update attempt: As suggested from Daniel Fischer's comment, a solution is presented in the answer section.

• Of course the assertion is wrong for finite $T_1$-spaces. If you have an infinite $T_1$-space $X$, consider a countable infinite subspace $Y$ of it. Find a subspace of $Y$ that satisfies the conclusion. If $Y$ doesn't have an infinite discrete subspace, then … – Daniel Fischer Jun 12 '17 at 14:01
• Sorry my bad! I forgot to add $X$ was an infinite set! So my aim should to first see for what condition there is a discrete sub space? And if there isn't then it's necessary to have a cofinite subspace homeomorphic to that given space? @DanielFischer – Mann Jun 12 '17 at 14:20
• Precise conditions for the existence of an infinite discrete subspace are hard. So the strategy is to show that if $X$ doesn't have a subspace of one type, then it must have a subspace of the other type. – Daniel Fischer Jun 12 '17 at 14:33
• I will try this approach! Thank you. – Mann Jun 12 '17 at 14:38
• I suppose you assume that $X$ is countably infinite? (Otherwise, pick a countable subspace and work with that.) There's a bit more to do. Starting with splitting $X$ into $A = \{ x \in X : \{x\} \text{ is open}\}$ and $S = X\setminus A$ isn't bad. If $A$ is infinite, we're done. Otherwise, there's more work to do, e.g. if $X \cong C \times D$ where $C$ is $\mathbb{N}$ with the cofinite topology and $D$ is discrete and contains at least two points, then $S = X$, but we need a proper subspace of $X$. Do you already know about connectedness? – Daniel Fischer Jun 12 '17 at 18:05

Suppose that (a countable space) $X$ contains no infinite cofinite subspace. We will show it has a countable discrete subspace.

So we start by finding a non-empty open subset $U_0$ of $X$ such that $X \setminus U_0$ is infinite, and we pick $x_0 \in U_0$.

Next we will choose $x_0, x_1,x_2, \ldots$ and open sets $U_0, U_1, U_2, \ldots$ by recursion such that when we have chosen $x_0, \ldots, x_{n-1}$ ,$U_0, \ldots U_{n-1}$ in such a way that for all $0 \le i,j \le n-1$:

• $x_i \in U_i$.

• $x_j \notin U_i$ for $j \neq i$.

• $A_{n-1} = X\setminus \bigcup\{U_m: 0\le m \le n-1\}$ is infinite.

Then we note that $A_{n-1}$ does not have the cofinite topology, so it has a relatively open non-empty subset $O$ with infinite complement in $A_{n-1}$, and so by $T_1$-ness (we have to avoid finitely many points) we have $U_n$ open in $X$ with $A_{n-1} \setminus (X \cap U_n)$ infinite and $U_n \cap \{x_0,\ldots,x_{n-1}\} = \emptyset$. This defines $X_n$ and finally we pick $x_n \in U_n \cap A_{n-1}$.

The last condition is needed to keep the recursion going and the first two show that the set $Y := \{x_n: n \in \mathbb{N}\}$ is an infinite discrete subspace of $X$ (as $U_n \cap Y = \{x_n\}$ for all $n$ and so all singletons are open in $Y$).

• Hi, I just wanted to ask is there any extension possible to my own answer, in a similar recursive way? And how did you approach this construction can you tell me what motivated you? I am not getting the intuition yet.I am guessing it has to do with removing open sets whose complement in infinite but by removing all such open set we must arrive at a finite cofinite topology which is just discrete space? There must not exist "so much" amount of open sets to be removed so i get a infinite cofinite space? – Mann Jun 13 '17 at 13:10
• @Mann so you want to assume no discrete subspace to get a cofinite one? The intuition for my construction is clear: pick an open set and a point in it, pick a second one that misses the first, a third one that misses the first two etc. In order to not let it end, we keep needing infinitely many fresh points, which is why we use non-cofiniteness. – Henno Brandsma Jun 13 '17 at 13:18
• I see I get your's now! But yes, i'd also like to have a hint for a construction of my case. If that's possibly. I think I was able to show that the space $(X,\tau)$ must have finitely many singleton open sets for a infinite discrete subspace to not exist. So I can get rid of these by removing finite amount of points. Then I have to prove that for the subspace i got, there must exist a infinite cofinite subspace of that subspace. – Mann Jun 13 '17 at 13:27
• The proof I gave is almost exactly that of the original paper. So I think it's pretty optimal. – Henno Brandsma Jun 14 '17 at 18:15
• @Andreo we use recursion not induction. It’s justified by the ZFC axioms. – Henno Brandsma Mar 3 '18 at 7:06

Lemma : If $$(X,\tau)$$ is a $$T_1$$ space and some finite set $$A=\left\{x_1,x_2,x_3, ..., x_n\right\}$$ is open in $$\tau$$ then the singleton sets $$\left\{x_i\right\}$$ such that $$i\in \left\{1,2,3,..,n\right\}$$ are also open in $$\tau$$.

We know that the set $$X\setminus A=X \setminus \cup^{n}_{i=1}\left\{x_i\right\}$$ is closed in $$\tau$$. Since $$(X,\tau)$$ is a $$T_1$$ space every singleton set is closed in $$\tau$$.

Finite union of closed set is closed. Hence, We know that $$\left(X \setminus \cup^{n}_{i=1}\left\{x_i\right\}\right)\cup \left\{x_k\right\} \cup \left\{x_l\right\} \text{ ... (n-1) times}$$ is closed where the index $$k\neq l \neq ...$$ .

Hence, we arrive at the conclusion that $$X\setminus \left\{x_i\right\}$$ is closed $$\implies$$ $$\left\{x_i\right\}$$ is open.

The contra-positive statement that if some finite amount singleton sets are not open in $$\tau$$ then any of their union is not open either.

Proof : I am going to assume that $$(X,\tau)$$ has no countably infinite subspace which is discrete and a space is not discrete iff $$\exists$$ $$\left\{x\right\}$$ for some $$x \in X$$ which is not open.

Now, If there were to be a finite amount of these sets, I could always remove these points to get a discrete space. Hence, there can't be a finite amount of these. Writing this condition more formally,

Denote the set $$A=\left\{x : \left\{x\right\} \text{ is not open in } \tau \right\}$$

Then, it is certain that $$X\setminus A$$ is finite. If it is not finite then either the space $$(X \setminus A, \tau_s)$$ can be countably infinite discrete space or one of its subspace is countably infinite discrete subspace.

Now consider the subspace $$(A,\tau_a)$$ which is obtained after removing finitely many singleton sets. By the lemma, this space can't have any finite open set and our only problem is now those open sets which are infinite and have an infinite complement as well.

Let $$\mathcal{O}=\left\{O\in \tau_a \; |\; A\setminus O \text{ is infinite.}\right\}$$

Since there are no finite open sets we know that for any $$O_i,O_j \in \tau_a$$ ,$$O_i\cap O_j$$ is either $$\phi$$ or another infinite $$O_k\in \mathcal{O}$$.

This implies that the set $$\mathcal{O}$$ can be partitioned by the equivalence relation: The sets $$\bigcap \mathcal{O_k}$$ are pairwise disjoint for each $$k \in K$$ and $$K$$ being some index set.

Picking one such $$\bigcap \mathcal{O_k}$$ and inducing a topology on it gets rid of any open set which has an infinite complement.

The resultant space is now either uncountably or countably infinite, whose subspace can be taken if needed.

This completes the proof that a cofinite subspace necessarily exist.