Take an infinite set $X$ and let $(X,\tau)$ be $T_0$ space which is not a $T_1$ space.
Then the space $(X,\tau)$ has a subspace homeomorphic to $(\mathbb{N},\tau_3)$, where $\tau_3$ is initial segment topology or the final segment topology.

I am unable to progress on this question with any particular idea. A few attempts i tried were:

1) Trying to guess and see if the initial segment and final segment topology are minimal $T_0$ spaces on Natural number. So if I can construct a minimal space by removing some points from $(X,\tau)$, i will be done.
2) I tried assuming one doesn't exist then other has to exist but i am not able to get explicit condition i.e., If there exist no subspace homeomorphic to say final segment topology then necessarily ... ?

Any particular hint will be really appreciated!

  • 1
    $\begingroup$ You mean, which has no infinite $T_1$ subspace... $\endgroup$ – Henno Brandsma Jun 17 '17 at 18:49

Suppose that $(X,\tau)$ is infinite and $T_0$, and suppose that it has no infinite $T_1$ subspace (which is what is needed for the five-space theorem).

intermediate motivational note. The five-space theorem (due to Sands and Ginsburg) says: consider $\mathbb{N}=\{0,1,2\ldots,\}$ and these five topologies:

  • $\tau_0 = \{\emptyset, X\}$, the indiscrete topology,

  • $\tau_1 = \{\emptyset\} \cup \{\{n \ge p\}, p \in \mathbb{N}\}$, the final segment topology,

  • $\tau_2 = \{\mathbb{N}, \emptyset\} \cup \{\{n: n \le p\}: p \in \mathbb{N}\}$, the initial segment topology,

  • $\tau_3 = \{\emptyset\} \cup \{X \setminus F: F \subseteq \mathbb{N} \text{ finite }\}$, the cofinite topology,

  • $\tau_4 = \mathscr{P}(\mathbb{N})$, the discrete topology.

Then if $X$ is an infinite topological space, it contains a subspace homeomorphic to one of these 5 spaces. This is minimal because each of these 5 spaces has the property that any infinite subspace of it is homeomorphic to the whole space. The proof of this goes along these lines: first rule out an infinite indiscrete space, and get an infinite subspace that is $T_0$. (see intermediate note 2 below) Then the proof in this post shows that either we have an infinite $T_1$ space (from an antichain), or an initial or final segment subspace. Finally, for the infinite $T_1$ subspace we show that either there is an infinite cofinite subspace, or there is a countable discrete subspace, which I showed in this question. Hence, that question and this one are the most interesting parts of the proof of this 5-space theorem. Hence the added assumption that $X$ has no infinite $T_1$ subspace. So an infinite $T_0$ space contains one of the last 4 spaces, an infinite $T_1$ topology one of the last 2, as the first 3 are not $T_1$, and $T_1$ is hereditary. And an infinite $T_2$ space, only contains a countable discrete subspace for that reason..

end of inserted note

Define $x \le y$ iff $x \in \overline{\{y\}}$ iff $\forall O \in \tau: (x \in O) \to (y \in O)$, the so-called specialisation pre-order. One checks that this is a pre-order: $x \le x$ for all $x$, is clear, and so is transitivity (consider the last reformulation which makes that clear).

For $T_0$-spaces it's a partial order: if $x \le y$ and $y \le x$, then $x = y$, for otherwise $x \neq y$ and the definition of $T_0$ says there is an open $O$ with $x \in O, y \notin O$ (which contradicts $x \le y$) or $x \notin O, y \in O$ (and this contradicts $y \le x$). So $x=y$ and we have a partial order.

inserted note 2

Note that $\le$ for general spaces $X$ only is a pre-order, i.e. reflexive and transitive, but not necessarily has the antisymmetric property that $x \le y \land y \le x \to x=y$; this property is in fact equivalent to the fact that $X$ is $T_0$. If we, in the proof of the five-spaces theorem, start with any infinite space $X$, we can consider this order, and define the standard equivalence relation $\sim$ induced by $\le$: $x \sim y$ iff $x \le y \land y \le x$, (note that being a preorder implies this is an equivalence relation) and note that we either have infinitely many different classes, or there is an infinite class of $\sim$. If the latter holds this class $A$ is a subspace with the indiscrete topology (any open set of it, if it contains one $x \in A$, it contains all of them) and we are done with the proof, or in the former case (infinitely many classes) pick one representative from each class and get an infinite $T_0$ subspace to continue the proof with.

end of inserted note 2

Now a standard fact in the theory of partial orders: any infinite partial order has a countable chain (a set of comparable elements) (up or down) or a countable antichain (a set of non-comparable elements). This follows e.g. from Ramsey's theorem and this and other proofs are here.

If an infinite antichain would exist, this would be an infinite $T_1$ subspace, as can easily be checked, so that options is ruled out (or we pass to that subspace and apply the cofinite vs discrete argument from before). So we have a countable descending chain $x_{n+1} \le x_{n}$ for all $n$ with maximum $x_0$ (all $x_n$ distinct in both cases), or a countable ascending chain $x_0$ as the minumum and $x_n \le x_{n+1}$ for all $n$.

In the last case, set $A = \{x_n :n =0,1,\ldots\}$ and suppose $O$ is open in $A$. Let $m = \min\{n: x_n \in O\}$, then $O \subseteq \{x_n: n \ge m\}$, but the reverse also holds (if $x_n \in A$, $n > m$ we have $x_m \le x_n$ and as $x_m \in O$, so also $x_n \in O$). Also, the sets $O_p:=\{x_n: n \ge p\}$ are all open in $A$ (this is clear for $p=0$ and if $p>0$ pick $O$ open with $x_p \in O, x_{p-1} \notin O$, as $x_{p} \not\le x_{p-1}$; we can do this to avoid all $x_l$ for $l < p$ (finite intersections) and then $p = \min\{n: x_n \in O\}$ and the above argument shows $O_p$ is indeed open. So open sets are essentially tails and all tails are open sets. So $x_n \to n$ is a homeomorphism of $A$ with $\mathbb{N}$ in the final segment topology.

If we have a descending chain, set $B =\{x_n, n=0,1,2\ldots\}$ and a similar argument shows that all open sets are initial segments and initial segments are open (use $\max$ instead of $\min$), so we have a homeomorphism of $B$ with $\mathbb{N}$ in the initial segment topology.

  • $\begingroup$ What is the 5-space theorem? $\endgroup$ – DanielWainfleet Jun 18 '17 at 1:13
  • $\begingroup$ imgur.com/a/79KaQ There is no mention of infinite $T_1$ subspace. $\endgroup$ – Mann Jun 18 '17 at 3:30
  • 1
    $\begingroup$ @DanielWainfleet every infinite topological space has a subspace homeomorphic to $\mathbb{N}$ in one of these 5 topologies: the indiscrete topology, the discrete topology ,the cofinite topology, the initial segment topologogy or the final segment topology. $\endgroup$ – Henno Brandsma Jun 18 '17 at 4:01
  • 1
    $\begingroup$ @Mann. It's a logical step in the proof to assume it, though. There need not be a final or initial segement subspace, if there only is an infinite antichain and no infinite chains. $\endgroup$ – Henno Brandsma Jun 18 '17 at 4:03
  • $\begingroup$ @DanielWainfleet I sketched the whole proof in this posting now. $\endgroup$ – Henno Brandsma Jun 18 '17 at 4:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.