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I am trying to understand part of a larger proof. I saw a statement that for a function $f:X\rightarrow\mathbb{R}$, where $X$ is compact and Hausdorff, $g((x,f(x))=x$ is continuous for all $x\in X$ because X is Hausdorff. Could someone explain why $X$ being Hausdorff is necessary and how the continuity of this projection follows from $X$ being Hausdorff?

HBHSU
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Define $\Gamma(f) = \{(x,f(x)) \in X \times \mathbb{R} : x \in X \}$, the graph of $f$. In the product topology on $X \times \mathbb{R}$ both projections are continuous by definition. Your map $g$ is just $\pi_X |_{\Gamma(f)}$ so always continuous, for any $f$, regardless of Hausdorffness of $X$.

Henno Brandsma
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  • Thank you for the answer... but then how is Hausdorffness necessary at all in this proof? https://math.stackexchange.com/questions/440676/graph-of-continuous-function-from-compact-space-is-compact – HBHSU Nov 26 '17 at 22:39
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    @HBHSU to see that we have a closed map: a compact set is closed in a Hausdorff space (but need not be in a non-Hausdorff space). I also gave an example where the statement fails for a non-Hausdorff space as well. – Henno Brandsma Nov 26 '17 at 22:43
  • I thought that any continuous function will map a closed set to a closed set, using page 126, Prop. 6.14 of Browder. – HBHSU Nov 26 '17 at 22:46
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    @HBHSU I don't have "Browder", but that's certainly not true. Such a map is called a closed map, and continuous maps need not be closed. – Henno Brandsma Nov 26 '17 at 22:48
  • I am confused about how to actually see that g maps closed sets to closed sets, then. – HBHSU Nov 26 '17 at 22:51
  • $f$ is defined on the graph. If $C$ is closed in the graph, it is compact (as a closed subset of the compact graph is also compact). The continuous $g$ sends compact sets to compact sets, and as in the image compact sets are closed by Hausdorffness, the image of $C$ is in fact closed. – Henno Brandsma Nov 26 '17 at 23:02