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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 $B=$
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 $F$

Let $xin E$ be the limit of a sequence $(x_n)_n$ where $x_nin F$ for all $n$. If $x=0$, nothing needs to be shown.
Once you know that $B:=,yin Fmid $ is compact, the continuous(!) map $BtoBbb R$, $ymapsto |y-x|$ attains its minimum at some $y_0in Bsubset F$. As $x_nto x$ 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 $x=y_0in F$.

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.

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?

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 $c_1,dots,c_nin K$ we have
$$|sum_i=1^nc_ie_i|geq rsum_i=1^n|c_i|.$$
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.

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 $F=textspane^1,cdots, e^n,$ where the $ langle e^i,e^jrangle =delta^ij; 1le ile n.$

Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots +a^i_ne^n; 1le ile n.$

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 $a^m_ito a_iin K; 1le ile n$.

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

Clearly, $xin F$ and now, another application of the triangle inequality shows that

$|x-x_n|to 0$, from which it follows that $x_nto xin F$

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.