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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.

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    $\begingroup$ 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 … $\endgroup$ Commented Jun 12, 2017 at 14:01
  • $\begingroup$ 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 $\endgroup$
    – Someone
    Commented Jun 12, 2017 at 14:20
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    $\begingroup$ 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. $\endgroup$ Commented Jun 12, 2017 at 14:33
  • $\begingroup$ I will try this approach! Thank you. $\endgroup$
    – Someone
    Commented Jun 12, 2017 at 14:38
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    $\begingroup$ 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? $\endgroup$ Commented Jun 12, 2017 at 18:05

1 Answer 1

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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 $Y$).

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  • $\begingroup$ 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? $\endgroup$
    – Someone
    Commented Jun 13, 2017 at 13:10
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    $\begingroup$ The proof I gave is almost exactly that of the original paper. So I think it's pretty optimal. $\endgroup$ Commented Jun 14, 2017 at 18:15
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    $\begingroup$ @Andreo we use recursion not induction. It’s justified by the ZFC axioms. $\endgroup$ Commented Mar 3, 2018 at 7:06
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    $\begingroup$ @jasmine Because we can intersect the first candidate $U_n$ we find with the (open!) complement of $\{x_0,\ldots,x_{n-1}\}$, to avoid the first finitely many points. We cannot say what $U_n$ will be, but certainly not what you see it as: why would that set even be open? And we're not picking it anyway if it were. $\endgroup$ Commented Jul 27, 2020 at 5:39
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    $\begingroup$ @jasmine That fact is trivial, because all $x_i$ are distinct. We want the larger set $U_n$ to be missing those points, and even all previous $U_i$... $\endgroup$ Commented Jul 27, 2020 at 5:57

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