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Suppose $S$ is any proper vector subspace of $\mathbb{R}^n$. Is $S$ a closed set in the usual topology on $\mathbb{R}^n$?

Geometrically, I think it is clear that $S$ must be closed in $\mathbb{R}^n$ for $1\leq n\leq 3$. If $x\in\mathbb{R}^n\setminus S$, then we can always find the shortest distance from $x$ to the point, line, or plane that is $S$ by taking the projection of $x$ onto $S$. Choosing $\epsilon$ to be less than this distance, $B_\epsilon(x)$ is an open ball around $x$ disjoint from $S$, so $S$ is closed.

I believe that the idea should carry over for higher dimensions, but I'm not sure how to make a more rigorous argument. How could this claim be proven without geometric handwaving? Thanks.

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3 Answers 3

up vote 3 down vote accepted

A vector subspace $V\subset\Bbb R^n$is always an intersection of hyperplanes (subspaces of codimension 1), $$ V=\bigcap_{i=1}^rH_i. $$ An hyperplane is closed, because it is the zero set of a linear form on $V$, which is continuous in the standard topology of $\Bbb R^n$. Thus $V$ is closed as intersection of closed sets.

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Let $u$ be the orthogonal projection of $x$ on $S$; then the Euclidean distance from $x$ to the nearest point of $S$ is $\|x-u\|$, so for any $\epsilon\le\|x-u\|$, the open $\epsilon$-ball centred at $x$ is disjoint from $S$.

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A bit more general: Linear maps between finite-dimensional spaces are always continuous.

Consequently, their kernels (invers image of zero) are closed. You can describe your subspace as kernel of a suitable linear map.

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