I got the Idea from this question: <> another thread

I got the idea to define a finite set like this:
We call a set $A$ finite if the topological space $(A,T)$ is hausdorff iff $T$ is the discrete topology. If my proof isn't wrong this one is equivalent to the normal defintion, that a set is finite if there is a bijection to $\{1,\dots,n\}$ for a $n\in \mathbb{N}$.

I never listen to any Topology-lecture (i didn't take a topology course) so i don't know if there is a problem in it, and I don't think I am able to do the proof on my own so I ask for some help.

My basic Idea was, that any bijection between discrete finite topological spaces is a homoeomorphism. Now we make a simplicial complex out of the sets (taking the elements of A as 0-skeleton and between two 0-cells a 1-cell) and perhabs we can show, that the homotopie groups aren't equal. (since the 0th is just a set we have to take a higher one). So for example, we can show that the free groups of $n$ and $n-1$ generators aren't the same.

Does anyone got an idea for this, or won't there be a way without cardinality at all.

  • $\begingroup$ yeah thats why the definitions works, it is compact only if we have finite sets ... $\endgroup$ – Dominic Michaelis Feb 19 '13 at 21:25
  • $\begingroup$ Ah, misread it. $\endgroup$ – Thomas Andrews Feb 19 '13 at 21:29
  • $\begingroup$ How are you constructing a simplicial complex from the set? That seems an odd leap, especially since you haven't defind finite, so it is hard to define a simplicial set. $\endgroup$ – Thomas Andrews Feb 19 '13 at 21:37
  • $\begingroup$ Oh right sry, my Idea was taking the elements of $A$ as $0$-skeleton and between two $0$-cells a 1-cell $\endgroup$ – Dominic Michaelis Feb 19 '13 at 21:41
  • $\begingroup$ downvoter why ? $\endgroup$ – Dominic Michaelis Apr 1 '13 at 11:03

I'm not completely sure what your question is. I think that at least part of what you are asking is: "Is it true that a set $X$ is finite $\iff$ the only Hausdorff topology on $X$ is the discrete topology?"

I didn't understand your proposal for how to prove this, but it is in fact true.

If $\tau$ is a Hausdorff topology on a finite set $X$, then since Hausdorff implies separated (or "$T_1$"), points are closed. Since finite unions of closed sets are closed, every subset is closed, hence every subset is open and the topology is discrete.

If $X$ is infinite, then there is an injection $\iota$ from the set $S = \{0\} \cup \{ \frac{1}{n} \ | \ n \in \mathbb{Z}^+\}$ to $X$. Let $X_1$ be the image of $\iota$ and let $X_2 = X \setminus X_1$. We topologize $X_1$ by saying that a subset $Y \subset X_1$ is open iff $\iota^{-1}(Y)$ is open in $S$, with its natural topology as a subset of $\mathbb{R}$. Thus $X_1$ is not discrete: $\{ \iota(0) \}$ is not open. On $X_2$ we put the discrete topology. Finally we endow $X$ with the topology it gets as the disjoint union "or coproduct" of $X_1$ and $X_2$. In this case, this amounts to saying that a subset $Y$ of $X$ is open iff $Y \cap X_1$ is open in $X_1$. This is a non-discrete Hausdorff topology on $X$.

  • $\begingroup$ I proofed it for myself over $f:X\rightarrow Y$ bijective with $X$ compakt and $Y$ hausdorff, and $f$ continuous, implies that $f^{-1}$ is continous, so we should get the result that there is only one topology in which it is hausdorff and compakt. My question is more about, if I can show the thing in the linked question, completly without a cardinality arguement $\endgroup$ – Dominic Michaelis Feb 19 '13 at 21:58
  • $\begingroup$ @Dominic: I'm not sure I understand your comment. Any infinite set carries infinitely many (non-homeomorphic) compact Hausdorff topologies. And I don't really get what you mean about showing that sets are infinite without a cardinality argument, although I may just be missing something here. $\endgroup$ – Pete L. Clark Feb 19 '13 at 22:01
  • $\begingroup$ sry i mean incomparable topologies (that on topology is a subset of another). I thought, that maybe with the definition above it is possible to show that there is no bijection between $X$ and $X\setminus\{0\}$ without using something like the pigeonhole principle or stuff like that $\endgroup$ – Dominic Michaelis Feb 19 '13 at 22:07
  • $\begingroup$ @Dominic: okay, it's true that any two compact Hausdorff topologies on a set are incomparable. (But I'm not sure how that is relevant to the present discussion.) I think I simply don't grok what it means to prove that a set is Dedekind finite without using notions like cardinality. (Asaf Karagila made a similar comment on the other thread, and he's a full time set theorist, so I would take his comments more seriously than mine.) $\endgroup$ – Pete L. Clark Feb 19 '13 at 22:12

Let $X$ be a topological space with at least three different points and $x \in X$. Suppose that the only Hausdorff topology on $X$ is the discrete one. If there exists a bijection $f : X \to X \backslash \{x\}$, then $f$ is a homeomorphism for discrete topologies. Now take the topology on $X$ defined by $\mathcal{T}=\{ O \subset X : O \subset (X\backslash \{x\}) \ \text{or} \ O \cap (X\backslash \{x\} )\neq \emptyset \}$. But $\mathcal{T}$ is a Hausdorff non discrete (since $\{x\} \notin \mathcal{T}$) topology on $X$: a contradiction.

  • $\begingroup$ Sorry but why do we need 3 different points ? $\endgroup$ – Dominic Michaelis Mar 30 '13 at 20:52
  • 1
    $\begingroup$ @DominicMichaelis: Otherwise $\mathcal{T}$ is not Hausdorff. $\endgroup$ – Seirios Apr 1 '13 at 16:51

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