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Let $\mathcal{S}$ be a linear subspace of the Euclidean space $\mathbb{R}^N$. Is $\mathcal{S}$ necessarily closed?

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closed as off-topic by mfl, Xander Henderson, Namaste, Shailesh, Parcly Taxel Aug 16 '18 at 4:32

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    $\begingroup$ Yes. Hint: Singletons are closed, and linear subspaces are preimages of singletons. $\endgroup$ – Tobias Kildetoft Aug 14 '18 at 17:32
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    $\begingroup$ What do you think? What have you tried? $\endgroup$ – mfl Aug 14 '18 at 17:36
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    $\begingroup$ Another basic (and rudimentary) approach is to start from the definitions, and use the fact that all norms are equivalent in finite-dimensional vector space in functional-analysis. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Aug 14 '18 at 17:40
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    $\begingroup$ Hint toward another approach: construct a basis for the subspace. Note the basis must be finite. Then consider possible limit points of sequences in the subspace. $\endgroup$ – Y. Forman Aug 14 '18 at 17:45
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    $\begingroup$ Then, according to Wiki for each $n\in \mathbb N$ there is one and only one (up to isomorphism [I guess homeomorphism]) topological vector space. It follows from this that any finite-dimensional vector subspace of a topological vector space is closed." $\endgroup$ – Dog_69 Aug 14 '18 at 17:49
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Indeed, vector subspaces are closed. Here's an explanation that you might find intuitive.

Let $P$ denote the (orthogonal) projection onto $\mathcal S^\perp$, the orthogonal complement of $\mathcal S$. Of course, $x$ is an element of $S$ if and only if $Px = 0$. Moreover, $P$ is a continuous map.

It follows that if $\{x_n\}$ is a sequence of points from $\mathcal S$ converging to $x$, then $Px_n$ must converge to $Px$. Since $Px_n = 0$ for all elements $x_n$, we must have $Px = 0$, and thus $x \in \mathcal S$.

So, whenever $x_n$ is a convergent sequence of points in $\mathcal S$ with limit $x$, we have $x \in \mathcal S$. Thus, $\mathcal S$ is a closed set.

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  • $\begingroup$ The terser version of this argument is what Tobias said, namely that $P^{-1}(\{0\})$ must be closed. $\endgroup$ – Omnomnomnom Aug 14 '18 at 17:48
  • $\begingroup$ Based on your proof, every linear subspace is closed! $\endgroup$ – Bashir Sadeghi Aug 14 '18 at 17:51
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    $\begingroup$ @BashirSadeghi indeed: in a finite dimensional normed vector space (such as a Euclidean space), every linear subspace is closed. In infinite dimensional spaces, the projection map $P$ might not be continuous or well-defined. $\endgroup$ – Omnomnomnom Aug 14 '18 at 17:53
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If you pick the basis correctly, the every linear subspace $W$ of $\mathbb R^N$ just looks like

$$W = \{ (x_1, ... , x_t,0, ... , 0) \in \mathbb R^n \}$$

which is a closed subset of $\mathbb R^N$.

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  • $\begingroup$ Such a change of basis will not generally preserve the Euclidean (i.e. inner-product) structure on $\Bbb R^N$, so a more flexible definition of the topology may be needed $\endgroup$ – Omnomnomnom Aug 14 '18 at 18:05

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