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Let $V$ be a vector space. Would you help me to prove that if $A$ is a subspace of $V$ then $A$ is convex and closed set.

I can prove that $A$ is convex (it's easy) and try to prove that $A$ is closed by showing that any sequences {$x_n$} in $A$ that converge to $x$ implies $x\in A$ but don't have any idea.


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Subspace of infinite dimensional normed linear spaces may not be closed. – William Sep 17 '12 at 23:14
Why? Could you give a counter example? – beginner Sep 17 '12 at 23:15
$l_\infty(\mathbb{R})$ with supremum norm. Let $M$ be the subspace of sequences with finitely many nonzero terms. Then the closure contains $(1, \frac{1}{2}, \frac{1}{3}, ...)$, which is not in $M$. – William Sep 17 '12 at 23:20
up vote 4 down vote accepted

I expanded my comments into an answer.

In infinite dimensional normed linear spaces, subspaces are convex but not necessarily closed.

Consider $l_\infty(\mathbb{R})$ which is the set of bounded sequences in $\mathbb{R}$ with the norm $|(a_n)_{n \in \omega}| = \sup a_n$. Note that the vector space structure is given by term by term addition and term scalar multiplication.

Then let $M$ be the subspace of sequences that have only finitely many nonzero terms. You can verify that $M$ is a subspace.

The sequence $\alpha = (1, \frac{1}{2}, \frac{1}{3}, ..., \frac{1}{n}, ...)$ is a bounded sequence, hence is in $l_\infty(\mathbb{R})$. $\alpha \in \overline{M}$, the closure, since

$|\alpha - (1, \frac{1}{2}, ..., \frac{1}{n}, 0, 0, 0, ...)|_\infty = \frac{1}{n + 1}$.

So $\alpha$ is the limit of a sequence of elements from $M$. $\alpha$ is in the closure of $M$ but not an element of $M$. Hence $M$ is not closed.

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In fact, in every infinite dimensional normed linear space that is complete (a.k.a Banach space), one can find a linear subspace that is not closed. – Theo Sep 18 '12 at 20:25

You mean $V$ is finite dimensional, no? We can then regard $V$ as the standard $\mathbb R^n$.

Consider a supplementary base $v_1,\ldots,v_n$ orthogonal to $A$. Then $$A=\{x \mid \forall i:\langle x,v_i \rangle =0\} = \bigcap_i {v_i}^\perp$$ and, as $\langle\cdot,v_i\rangle$ is continuous, we have that ${v_i}^\perp$ is closed.

For infinite dimension, this is not true. Neither the topology is not uniquely determined, can be of many kind..

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I suppose $V$ is a finite dimensional vector space over $\mathbb{R}$. We can identify $V$ as $\mathbb{R}^n$. Let $A$ be an $n - r$ dimensional subspace of $V$. There is a bijective linear transformation $f\colon V \rightarrow V$ such that $f(A) = \{(x_1\dots,x_n) \in \mathbb{R}^n\colon x_1 =\cdots=x_r = 0\}$. Hence $A$ is closed.

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