If $X$ is a normed space and $Y$ a finite dimensional subspace, then $Y$ is closed.

Question

I am trying to produce a direct proof on the statement mentioned above. The field I am working in is $\mathbb{R}$. My proof outline goes as following:

If $Y$ is finite-dimensional, there exists a basis $(e_1,\dots,e_n)$ with $n<\infty$ such that every $y\in Y$ can be written as $y = \sum_{i=1}^n a_i e_i$ with $(a_1,\dots,a_n)$ constants in $\mathbb{R}$. So I need to take a sequence $(y_n)_{n\in\mathbb{N}}$ in Y that converges to some element in $X$.

Here is where my argument is fuzzy. As $X$ is not necessarily finite dimensional, what I understand is that if we can think of the sequence as converging to some $(x_1,x_2,\dots, x_{n-1},x_n,0,0,0,0,\dots)$, is this correct?

I am stuck at this and I do not know how to proceed. Any ideas would be appreciated.

Answer 1

Assume that $\{e_1,\ldots,e_k\}$ is a basis of $Y$. Let $$ d=\inf_{\lvert c_1\rvert+\cdots+\lvert c_k\rvert=1}\|c_1e_1+\cdots+c_ke_k\|. \tag{1} $$ Clearly, $d>0$, since the set $K=\{(c_1,\ldots,c_k):\lvert c_1\rvert+\cdots+\lvert c_k\rvert=1\}$ is compact, and $f(c_1,\ldots,c_k)= \|c_1e_1+\cdots+c_ke_k\|$ is continuous and non-vanishing on $K$.

$(1)$ provides the following inequality: $$ \|c_1e_1+\cdots+c_ke_k\|\ge d\big(\lvert c_1\rvert+\cdots+\lvert c_k\rvert\big), $$ now for all $c_1,\ldots,c_k\in\mathbb R$ (or $\mathbb C$).

Assume now that $\{y_n\}$ is a Cauchy sequence in $Y$, and $y_n=c_{n,1}e_1+\cdots+c_{n,k}e_k$. Then for every $\varepsilon>0$, there exists an $n_0$, such that $m,n\ge n_0$ implies that $$ \varepsilon>\|y_m-y_n\|=\|(c_{m,1}-c_{n,1})e_1+\cdots+(c_{m,k}-c_{n,k})e_k\|\ge d\big(\lvert c_{m,1}-c_{n,1}\rvert+\cdots+\lvert c_{m,k}-c_{n,k}\rvert\big), $$ which implies that all the sequences $\{c_{n,k}\}_{n\in\mathbb N}$ are Cauchy, and hence $c_{n,k}\to c_k$, and thus $$ y_n\to y=c_1e_1+\cdots +c_ke_k\in Y. $$ Indeed $Y$ is closed.

Answer 2

Hint: If $x_n\rightarrow x$, then you reduce to the finite dimensional case by adding $x$ to the basis and using the inherited norm.

Answer 3

Based on your statement of the problem, you don't assume that $X$ is closed. So I don't think it is clear on the front end that the limit of a sequence is an element of $X$.

Instead here is a hint. Assume $(y_k)_{k=1}^\infty$ is a cauchy sequence. Express each $y_k = a_{1,k}e_1 + \cdots a_{n,k}e_n$ using your basis. Use the definition of cauchy to prove that for each $i$, the sequence $(a_{i,k})$ is a cauchy sequence of real numbers, which then must have a limit $a_i$. Then prove that $y_k \to a_1 e_1 + \cdots + a_n e_n$.