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This was a bonus question on a recent homework in my point-set topology class. I've since turned in the assignment but still don't have an answer to this question.

Let $X$ be a locally compact Hausdorff topological space. Let $X^+$ denote the one-point compactification of $X$. Prove or find a counterexample to the following claim: If $X^+$ is connected, then $X$ has no connected components which are compact.

The converse is true and straightforward to show. If we add the assumption that $X$ is locally connected, then the statement is true. And if we remove the requirement of local compactness, then it's false. But in its current form I haven't been able to make heads or tails of it.

Silvio Mayolo
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  • The problem should probably say "which are compact and different from $X$". Assume that $K\subsetneq X$ is a compact connected component of $X$. Then $K\coprod (X\setminus K)^+$ is $X^+$, which would be disjoint. So, you only need to show that $K\coprod (X\setminus K)^+$ is the one-point compactification of $X$. It is compact as the disjoint union of two compacts. $X$ is homeomorphic to it minus the extra point. – conditionalMethod Oct 25 '19 at 21:00
  • @conditionalMethod Why not an official answer? – Paul Frost Oct 25 '19 at 22:36

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