Let $(x_i) _{i \in I}$ be a net in a compact Hausdorff space X, with the property that every convergent subnet has the same limit $x$. Does this imply that $(x_i) _{i \in I}$ converges?

If I try to proceed by contradiction, I do not how to rule out the possibility of having subnets that do not converge at all. (Of course, by compactness each such subnet should have at least a subnet convergent to $x$, but this does not help me.)

(If it helps, my compact space is in fact a weakly-compact subset in a locally-convex linear topological space space, obtained with Alaoglu's theorem.)


Suppose, toward a contradiction, that your net $(x_i)_{i\in I}$ doesn't converge to $x$. So $x$ has an open neighborhood $N$ such that the net doesn't ultimately get into $N$. That is, if you define $J=\{i\in I:x_i\notin N\}$, then every element of $I$ is $\leq$ an element of $J$. Use that to check that $(x_i)_{i\in J}$ is a subnet of your original net $(x_i)_{i\in I}$. By compactness (of $X-N$, which is compact because $X$ is compact and $N$ is open), some subnet of $(x_i)_{i\in J}$ must converge to some point $y\in X-N$. But that subnet is also a subnet of your original net $(x_i)_{i\in I}$, so it's not allowed to converge to anything other than $x$ --- contradiction.

  • $\begingroup$ I have just finished the proof under the stronger assumption that each element of $I$ has at least one successor. It follows essentially the same line of thought as yours. Before accepting your answer, let me first check that $J$ is upper-directed, in order to make sure that $(x_j) _{j \in J}$ is indeed a subnet. (Is it obvious? It seems to me that it isn't directed.) $\endgroup$
    – Alex M.
    Sep 18 '15 at 16:43
  • 1
    $\begingroup$ To check that $J$ is directed, consider any two elements of $J$. They have an upper bound in $I$, because $I$ is directed. But that element of $I$ (like every element of $I$) has an element of $J$ above it, and that element of $J$ serves as the upper bound you need in $J$. $\endgroup$ Sep 18 '15 at 16:48

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