Are Normed linear spaces $T_4$(Normal Hausdorff)?

Question

Do normed linear spaces have the properties of a normal hausdorff space?

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I just sat an exam and I couldn't work out how to prove something initially, then I assumed that normed linear spaces are $T_4$, and the proof was easy. Are they? Please say yes :)

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The proof for the record was showing that if we have some normed linear space $X$, and the unit ball on this space has a compact surface, prove that $X$ is not infinite dimensional. Which after the exam I realized was a consequence of Riesz's lemma. But without that I took an open covering and using my $T_4$ properties there were disjoint open balls covering every element of the compact surface, which would cause contradiction for my finite subcovering over infinite elements.

Answer 1

Every metric space is Hausdorff and $T_4$: Hausdorffness is easy to check. For $T_4$, we use the reciprocal of Urysohn's Lemma: take $F,K \subset X$ be disjoint closed sets. Then: $$f: X \to \Bbb R \\ f(x) = \frac{{\rm d}(x,F)}{{\rm d}(x,F)+{\rm d}(x,K)}$$ is well-defined, continuous, and $f^{-1}([0,1/2[) \supset F$, $f^{-1}(]1/2,1]) \supset K$ are disjoint open sets (because $[0,1/2[$ and $]1/2,1]$ are open in $[0,1]$).

Normed linear spaces are particular metric spaces, so yes, normed linear spaces are Hausdorff and $T_4$.