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Nov 28, 2016 at 18:57 comment added Alexander Körschgen Thank you for your comment. Unfortunately, the space $X$ is not weak Hausdorff. If we endow $K := \{c,d\}$ with the discrete topology, $K$ is compact Hausdorff. However, the map $K \to X$ given by $c \mapsto a, d \mapsto x$ is continuous while its image $\{a,x\}$ is not closed in $X$.
Nov 26, 2016 at 3:06 history answered Jeff Strom CC BY-SA 3.0