Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks.

share|improve this question
1  
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
1  
$l_\infty(\mathbb{R})$ 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
add comment

3 Answers

up vote 3 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.

share|improve this answer
    
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
add comment

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..

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.