why a subspace is closed?

Question

Let $E$ be $K-$vector space with a norm $\|\cdot\|$, and $F$ a subspace with dimension $n$. Show that $F$ is a closed set .

I am trying to show that any convergent sequence of elements $(x_n)_{n \in N}$ of $F$ converge to an element $x \in F$.

This means I need to show that $||x_n-x|| \rightarrow 0$

I see that like a distance which is attended by the fuction $y \rightarrow ||x-y||$ .

How can I show that $ x \in F$?

I have been reading a solution which I don't really understand the intuition behind it :

let $x$ be in $E$ and let $B=\{ y \in F ||y-x|| \leq ||x||\}$ after showing that its a compact, they tried to show that $inf_{y \in F}(||x-y||)= \lambda $ exists.

by definition of $\lambda$ it is the distance between $x$ and $F$

Answer 1

Let $x\in E$ be the limit of a sequence $(x_n)_n$ where $x_n\in F$ for all $n$. If $x=0$, nothing needs to be shown. Once you know that $B:=\{\,y\in F\mid \|y-x\|\le \|x\|\,\}$ is compact, the continuous(!) map $B\to\Bbb R$, $y\mapsto \|y-x\|$ attains its minimum at some $y_0\in B\subset F$. As $x_n\to x$ and $\|x\|>0$, almost all $x_n$ are $\in B$. Then from $\|y_0-x\|\le \|x_n-x\|\to 0$, we conclude $\|y_0-x\|=0$ and finally $x=y_0\in F$.

Answer 2

You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.

But I think it may be easier to show that the complement of F is open. Consider an element $x \in E \setminus F$. It has a distance $\gamma \gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y \in F$ and $z \perp F$. Then $B(x, \gamma /2) \subseteq E \setminus F$ so $x$ is in the interior of $E \setminus F$. We chose $x$ arbitrarily, so this proves $E \setminus F$ is open so that $F$ is closed.

Answer 3

Another approach will be to show that a finite-dimensional normed space (over $\mathbb R$ or $\mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.

Answer 4

Hint:

1. Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.

2. Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.

3. Find a subsequence which has a limit in $F$. What is the limit?

Answer 5

To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis $\{e_1,\dots,e_n\}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,\dots,c_n\in K$ we have $$\|\sum_{i=1}^nc_ie_i\|\geq r\sum_{i=1}^n|c_i|.$$ Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)\subset F$. We know that each $x_m=\sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.

Answer 6

I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:

Let $F=\text{span}\{e^1,\cdots, e^n\},$ where the $ \langle e^i,e^j\rangle =\delta^{ij};\ 1\le i\le n.$

Now, suppose $F\supset (x_n)\to x\in E$. Then, $x_i=a^i_1e^1+\cdots +a^i_ne^n;\ 1\le i\le n.$

Of course, $(x_n)$ is a Cauchy sequence in $E$, so since

$\|x_m-x_k\|^2=\|(a^m_1-a^k_1)e^1+\cdots (a^m_n-a^k_n)e^n\|^2=|a^m_1-a^k_1|^2+\cdots +|a^m_n-a^k_n|^2$

(because the $(e^i)$ are orthogonal), each $(a_i^m)$ is a Cauchy sequence in $K$.

Assuming now that $K$ is complete, we have $a^m_i\to a_i\in K;\ 1\le i\le n$.

To finish, consider the vector $x=a_1e^1+\cdots +a_ne^n$.

Clearly, $x\in F$ and now, another application of the triangle inequality shows that

$\|x-x_n\|\to 0$, from which it follows that $x_n\to x\in F$