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To prove that finitely generated spaces are corecompact, I used the following characterisation of finitely generated topological spaces:

In a finitely generated topological space every point has a smallest neighbourhood.

Now I was wondering how to prove this characterisation. The definition I use for finitely generated topological spaces is the following:

$\forall A \in \mathcal{P}(X), \forall x \in X: x \in cl(A) \Rightarrow \exists a \in A, x \in cl\{a\}$

I tried proving it by contradiction: Assume that there is no such thing as a smallest neighbourhood. Suppose that $\forall V \in \mathcal{V}(x) \exists W \in \mathcal{V}(x); W \subseteq V$. Can someone give me a hint so I can come to a contradiction? Or is there a nicer way to prove this?

As always, any hints or comments will be appreciated!

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I personally found your definition of a finite generated topological space hard to comprehend, so perhaps others may as well. For our benefit, I digested it down to this (please correct me if I am wrong): If $A$ is a subset of $X$ and $x\in X$, then $x\in \bar{A}$ implies that $A \cap U \neq \emptyset $ for all open neighbourhoods $U$ of $x$. –  Ragib Zaman May 12 '12 at 12:34
    
And now that I've written that out, it appears that is nothing more than what it means for $x$ to be in the closure of $A.$ –  Ragib Zaman May 12 '12 at 12:39
    
@RagibZaman: There is no point $x$ in $(0,1)$ such that $0$ is in the closure of $\{x\}$. –  Michael Greinecker May 12 '12 at 12:43
    
@Michael Greinecker, Thanks. Could you please point out my error in this then?: 1) By definition of closure, $x$ in the closure of $\{a\}$ means every neighbourhood U of $x$ has non-empty intersection with $\{ a \}.$ 2) U has non-empty intersection with $ \{ a \} $ if and only if $a \in U.$ 3) To say "There exists $a \in A$ such that $x \in \text{cl} \{ a \} $ is thus equivalent to " $ A \cap U $ is non-empty". –  Ragib Zaman May 12 '12 at 12:50
    
@RagibZaman In (3) above, the element of $A \cap U$ may vary with $U$. For $x \in \mathrm{cl}\{a\}$ we need $a \in U$ for every $U$. –  martini May 12 '12 at 12:54

2 Answers 2

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Let $U$ be the intersection of all neighborhoods of $x$. We show that $U$ is open. Let $y$ be in the closure of $X\backslash U$. We have to show that $y\in X\backslash U$. By assumption, there is $z\in X\backslash U$ such that $y$ is in the closure of $\{z\}$. Now by construction of $U$, there is an open neighborhood $V$ of $x$ such that $z\notin V$. Since $V$ is open, the closure of $\{z\}$ is not in $V$ and hence $y\notin V$. Hence $y\notin U$ and $y\in X\backslash U$.

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This proof method also proves the stronger statement that in finite generated spaces, arbitrary intersections of open sets are open. –  Ragib Zaman May 12 '12 at 13:08

If you show that, in a finitely generated space, any intersection of open sets is open, then it is easy to see that every point $x$ has a smallest neighborhood - it is the intersection of all open sets containing $x$.

In fact, the claim that intersection of any system of open sets is open is a characterization of finitely-generated spaces, see Wikipedia.

However, let us have a look only at the direction important for you: You want to show that any finitely-generated space (according to your definition) has this property. This is equivalent to showing that arbitrary union of closed sets is closed.

Let $F=\bigcup_{i\in I}F_i$. Let $x\in\overline{F}$, then there is an $a\in F$ such that $x\in\overline{\{a\}}$. This implies the existence of an $i\in I$ such that $x\in\overline F_i=F_i$ (we take $i\in I$ for which $a\in F_i$). Clearly, this implies $x\in F$. We have shown that $F$ is closed.

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