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I am trying to prove the following statement

If $(x_{\alpha})_{\alpha \in \Lambda}$ verifies that every sub-net has a sub-sub-net which converges to $x$, then $(x_{\alpha})_{\alpha \in \Gamma}$ converges to $x$.

I think it would be useful to let you know the definition of sub-net I am using:

$\{y_e\}_{e \in E}$ is a subnet of $\{x_d\}_{d \in D}$ (where $D$ and $E$ are directed (filtered) sets) if and only if there is a function $f: E \to D$ that verifies the following two properties:

  1. $\forall d \in D$ there is $e' \in E$ such that for all $e\geq e'$, $f(e) \geq d$.

2.$\forall e \in E$, $y_e=x_{f(e)}$

Now, I've tried to show the statement straightforward but I couldn't. So maybe it is easier to show it by the absurd:

Suppose $(x_{\alpha})_{\alpha \in \Lambda}$ doesn't converge to $x$. Then, there is some $U$ open with $x \in U$ such that for all $\alpha$, there is $\alpha'$ with $\alpha'\geq \alpha$ such that $x_{\alpha'} \not \in U$.

I don't know how to get a contradiction from these affirmations. I would appreciate suggestions.

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You have already suggested yourself to start by assuming that $(x_\alpha)_{\alpha\in\Lambda}$ does not converge to $x$.

You wrote:

Suppose $(x_{\alpha})_{\alpha \in \Lambda}$ doesn't converge to $x$. Then, there is some $U$ open with $x \in U$ such that for all $\alpha$, there is $\alpha'$ with $\alpha'\geq \alpha$ such that $x_{\alpha'} \not \in U$.

Let us fix one such neighborhood $U$ and rewrite this slightly: For every $\alpha\in\Lambda$ let us choose $\beta(\alpha)\in\Lambda$ with the property that $\beta(\alpha)\ge\alpha$ and $x_{\beta(\alpha)}\notin U$. (This is basically the same thing you wrote, but I wanted to stress that this element depends on $\alpha$ and also to choose exactly one such element for each $\alpha$.)

Let us denote $D=\{\beta(\alpha); \alpha\in\Lambda\}$. It should be relatively easy to show that $D\subseteq\Lambda$ is a directed set and $(x_\alpha)_{\alpha\in D}$ is a subnet of $(x_\alpha)_{\alpha\in\Lambda}$. Since all elements of $(x_\alpha)_{\alpha\in D}$ are outside the open set $U$, no subnet of this net can converge to $x$.


EDIT: Here follows my original answer, which is unnecessary complicated; but I decided to keep it, some of the facts mentioned there might be useful for someone reading this.

This only works for Tychonoff spaces, since I am using compactification of $X$. I hope this is not too restrictive for you. (But I suppose someone will post proof without this assumption.)


Let us consider the same net $(x_\alpha)_{\alpha\in\Lambda}$ in some compactification $cX$ of $X$.

This make things easier: Working in a compact Hausdorff space, we know that every net has a cluster point. We also know that a net converges if and only if it has unique cluster point. Another useful fact is that $y$ is a cluster point of some net if and only if there is a subnet converging to $y$.

Now if $(x_\alpha)_{\alpha\in\Lambda}$ does not converge to $x$, then there is a cluster point $y$ of $(x_\alpha)_{\alpha\in\Lambda}$ such that $y\ne x$. (Otherwise $x$ would be unique cluster point, hence it would be the limit.)

So there is a subnet $(x_\alpha)_{\alpha\in\Lambda'}$ which converges to $y$ in $cX$. Then every subnet of $(x_\alpha)_{\alpha\in\Lambda'}$ also converges to $y$ in $cX$. Clearly, such subnet does not converge to $x$ in $X$ (or in $cX$, which is the same thing).

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  • $\begingroup$ Thanks for the answer, though I haven't seen Tychonoff spaces, I am supposed to prove it for a general topological space. If you can think of an answer or a hint that helps me to show it for $X$ arbitrary topological space, then you can add it to your previous answer $\endgroup$ – user100106 Aug 15 '14 at 14:06
  • $\begingroup$ I have edited my post. $\endgroup$ – Martin Sleziak Aug 15 '14 at 14:16

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