# Subspaces of finite dimensional Hilbert spaces

This might be a trivial question but please point out exactly where my reasoning is incorrect. Is every subspace of $\mathbb{R}^n$ closed since $\mathbb{R}^n$ with the dot product is a finite dimensional Hilbert space and all subspaces of finite dimensional Hilbert spaces are closed?

• It's not incorrect, but you need much less than Hilbert. Every finite-dimensional subspace of a Hausdorff topological vector space is closed. – Daniel Fischer Feb 16 '14 at 16:21
• Do you need the TVS to be Hausdorff? Does it not follow for all topological vector spaces? – Alex Feb 16 '14 at 16:44
• @Alex Hausdorff is necessary. Otherwise $\{0\}$ is a finite-dimensional but not closed subspace. – Daniel Fischer Feb 16 '14 at 17:56
• I don't understand your '$\{0\}$ example' above, could you expand on that? Why would $\{0\}$ not be closed if assuming that TVS is the only requirement and why is that a problem? – Alex Feb 18 '14 at 8:43
• @Alex Let $X$ be the real line with the trivial topology: the only open sets are $\varnothing$ and $X$. This is a TVS. The set $\{0\}$ is a zero-dimensional subspace of $X$ which is not closed. If you don't like zero-dimensional example, consider $\mathbb R$ as a subspace of $\mathbb R^2$, again with the trivial topology. – user127096 Feb 19 '14 at 2:23

## 1 Answer

The statement in the comment is correct.

The reasoning in the question is also correct on the technical level, but I can't really tell how adequate it is in the context where you'll put it. Generally, by the time people learn what a Hilbert space is, they already know how to prove that a linear subspace of $\mathbb R^n$ is closed (because they know enough linear algebra to represent such a subspace by a system of linear equations, i.e., as the kernel of a linear operator).

What I'm saying is: a person who actually needs an explanation of why a linear subspace of $\mathbb R^n$ is closed is unlikely to be satisfied with yours.

• Okay I have not followed traditional course of studying functional analysis but am instead trying to learn independently. Okay but just to be clear my question was referring to any subset of $\mathbb{R}^{n}$ which is endowed with the subspace topology, not just linear subspaces of $\mathbb{R}^{n}$. Does the result I stated in my initial question only apply to linear subspaces? – user103184 Feb 23 '14 at 10:55
• @Moses So you're asking if every subset of $\mathbb R^n$ is closed? Does the interval $(0,1)$ look like a closed subset of $\mathbb R$ to you? – user127096 Feb 23 '14 at 19:12