Suppose $E$ a topological vector space,which need not be Hausdoff. $x,y\in E$ are different.
How to prove the close line segment $\{\alpha x+(1-\alpha)y:\alpha\in[0,1]\}$ is closed. And should it be compact?
2026-03-27 02:39:50.1774579190
A question about close line segment in TVS.
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It will be compact, as the function that sends $\alpha \in [0,1]$ to $\alpha x + (1-\alpha)y$ is continuous (as all vector operations are, details omitted). And the continuous image of a compact space ($[0,1]$) is compact.
Compact need not imply closed for non-Hausdorff spaces. To give a rather extreme counterexample, suppose that $E$ has the indiscrete (a.k.a. trivial) topology: only $E$ and $\emptyset$ are open. It's quite clear (check this!) that this makes $E$ into a topological vector space, and a very non-Hausdorff one at that. The closed line segment is then not closed, as no non-empty proper subset of $E$ is.