Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
yeah thats why the definitions works, it is compact only if we have finite sets ... –  Dominic Michaelis Feb 19 '13 at 21:25
    
Ah, misread it. –  Thomas Andrews Feb 19 '13 at 21:29
    
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. –  Thomas Andrews Feb 19 '13 at 21:37
    
Oh right sry, my Idea was taking the elements of $A$ as $0$-skeleton and between two $0$-cells a 1-cell –  Dominic Michaelis Feb 19 '13 at 21:41
    
downvoter why ? –  Dominic Michaelis Apr 1 '13 at 11:03
show 1 more comment

2 Answers

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 $X$.

share|improve this answer
    
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 –  Dominic Michaelis Feb 19 '13 at 21:58
    
@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. –  Pete L. Clark Feb 19 '13 at 22:01
    
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 –  Dominic Michaelis Feb 19 '13 at 22:07
    
@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.) –  Pete L. Clark Feb 19 '13 at 22:12
add comment

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.

share|improve this answer
    
Sorry but why do we need 3 different points ? –  Dominic Michaelis Mar 30 '13 at 20:52
1  
@DominicMichaelis: Otherwise $\mathcal{T}$ is not Hausdorff. –  Seirios Apr 1 '13 at 16:51
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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