Are linear subspaces of euclidian spaces closed

Question

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

Answer 1

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.

Answer 2

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 $\mathbb R^N$.