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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.

Someone
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    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 – Someone Jun 12 '17 at 14:20
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    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. – Someone Jun 12 '17 at 14:38
  • @DanielFischer , If i assume that $(X,\tau)$ has no infinite discrete space then it has to be the case that there are going to at least so many singleton set which are not open, Denote that by some set $S=\left{x\in X : \left{x\right} \text{ is not open }\right}$ in $(X,\tau)$, so that $X\setminus S $ is finite. Then i can remove all the points of the set $X \setminus S$ from the space $(X,\tau)$ is this approach correct? – Someone Jun 12 '17 at 17:39
  • Will this be sufficient, If i show that either it is a discrete space and if it is not then it is a co-finite topology? – Someone Jun 12 '17 at 17:44
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    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
  • @DanielFischer, does the proof look okay now? – Someone Jun 16 '17 at 18:31
  • @Mann, it looks like you are trying to use an equivalence relation and split $\mathcal{O}$ into equivalence classes. I can't really understand what kind of equivalence relation you are using ... open sets $O_i$ and $O_j$ are equal if ... ? – Andreo Feb 28 '18 at 23:12
  • Possible duplicate of Let $X$ an infinite $T_1$ space, then exist some subspace homeomorphic to $(\Bbb N,\tau)$ where $\tau$ is discrete or cofinite – Cassius12 Nov 30 '18 at 15:26

1 Answers1

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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 $Y$).

Henno Brandsma
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  • 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? – Someone 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. – Someone Jun 13 '17 at 13:27
  • My idea was to do something like this, I know that since there's no finite open set, every infinite open set whose complement is infinite say $O_1$ and $O_2$ their intersection must be empty or infinite as well. If pick the "smallest" such open such in a kind,for example in the above i can pick $O_1 \cap O_2$ if only $O_1$ and $O_2$ existed, I can just take that set to be my subspace. – Someone Jun 13 '17 at 13:30
  • This problem seems to be part of the "5-spaces theorem": every infinite topologic space has a subspace homeomorphic to $\mathbb{N}$ in one of these 5 topologies: the indiscrete topology, the "up" topology (with all tails open), the "down" topology (with all initial segments open), the cofinite topology or the discrete topology. The proof by Ginsburg and Sands (in "minimal infinite topological spaces") has this problem as the final part of its proof. The first three spaces can be ruled out for $T_1$ spaces, and can be reached via considerations about the specialisation order. – Henno Brandsma Jun 14 '17 at 03:46
  • Yes, that's exactly what it is! Actually I am following a book called "topology without tears" and there, the same problem has been divided into 5 subpart to prove. This was one of the subpart. The next subpart is indeed the initial segment topolgy and final segment topolgy where it is given that infinite space is a $T_0$ space but not $T_1$ space. – Someone Jun 14 '17 at 06:12
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    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
  • @HennoBrandsma, could you please explain your proof?

    a) I don't understand this part: "and so by $T_1$-ness (we have to avoid finitely many points)"

    b) Is $O = A_{n−1} \cap U_n$ or $X_n = A_{n−1} \cap U_n$?

    c) Why is a singleton set ${x_n}$ open?

     – Andreo Mar 01 '18 at 01:46
  • @Andreo We have $A_{n-1}$ defined at stage $n$ (as the complement of all previous $U_i$) and we have $O$ open in $X$ such that $O \cap A_{n-1}$ is open and $O \cap A_{n-1}$ is not cofinite. We take $U_n = O \setminus {x_0,\ldots,x_{n-1}}$ which is open by $T_1$-ness (we avoid the previous finite set which is closed) and pick $x_n \in U_n$. The new set $A_{n+1}$ is then infinite because $O$ has infinite in $A_{n-1}$. The last part I addressed in an edit. – Henno Brandsma Mar 01 '18 at 07:58
  • @HennoBrandsma, thank you for the clarification. Now it is clear. After this proof I'm thinking about induction. Why can we use induction in this proof to build an infinite discrete subspace? However we can't use induction in other cases, for instance, to build an infinite intersection of open sets. We could say that, by induction, an infinite intersection of open sets is an open. But it is not true in general. – Andreo Mar 02 '18 at 23:39
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    @Andreo we use recursion not induction. It’s justified by the ZFC axioms. – Henno Brandsma Mar 03 '18 at 07:06
  • @HennoBrandsma sir how $U_n \cap {x_0,\ldots,x_{n-1}} = \emptyset$? u said in above that $x_i \in U_i$ My thinking is that $U_n={x_0,\ldots,x_{n}}$ and now ${x_0,\ldots,x_{n}} \cap {x_0,\ldots,x_{n-1}}={x_0,\ldots,x_{n-1}} \neq \emptyset$ – jasmine Jul 27 '20 at 04:47
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    @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. – Henno Brandsma Jul 27 '20 at 05:39
  • okss sir thanks u @Henno that mean u want to say that${x_n} \cap {x_0,x_1,x_2,......x_{n-1}} = \emptyset$ – jasmine Jul 27 '20 at 05:55
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    @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 $U_i$... – Henno Brandsma Jul 27 '20 at 05:57