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I have been looking around for how to prove that a finite dimensional subspace of a normed vector space is closed. I've seen it be mentioned a few times that you can use equivalence of norms but I haven't been able to find the actual proof. I'm guessing you start off by expanding your convergence sequence in terms of basis vectors and somehow use the equivalent norm that pulls out the coefficients.But I'm unsure on the details, where you would go from there and even if this is the correct approach.

Any help is appreciated, cheers.

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  • $\begingroup$ There's a proof here. $\endgroup$
    – user771918
    Apr 28, 2020 at 2:34

2 Answers 2

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Take the normsup, $\|x\|=sup(|x_1|,...,|x_n|)$ to show it is closed, Let $F$ be a finite dimensional subspace, if $u_n\in F$ converges towards $u$, $(u_n)$ is a Cauchy sequence, so for the normsup, and the coordinates of $(u_n)$ are Cauchy sequences for the normsup and converges towards $u^1,...,u^n$, this implies that $u_n$ converges in towards $u$ for the normsup and for the the restriction of the original norm to $F$ (since it is equivalent to the normsup), since limit are unique, $(u_n)$ converges towards $u$.

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Let $b_1,...,b_n$ be a basis for the finite dimensional space $F$. Define $c_j$ by $\sum_j c_j(y) b_j = y$. The $c_j$ are continuous linear functionals on $F$.

Define $n(x) = \|\sum_k x_k b_k \|$ and note that $n$ is a norm on $\mathbb{R}^n$ and is equivalent to $\| \cdot \|_1$.

To see that the $ c_j$ are continuous, note that \begin{eqnarray} |c_j(y)| &=& |c_j(\sum_k \alpha_k b_k)| = |\sum_k \alpha_k c_j(b_k)| \le \sum_k |\alpha_k|(\max_k |c_j(b_k)|) \\ &=& K\|\alpha\|_1 \le K'n(x) = K' \| \sum_k \alpha_k b_k \| = K\|y\| \end{eqnarray}

Suppose $y_k \to y$ with $y_k \in F$. Then \begin{eqnarray} \|y-\sum_j c_j(y)b_j \| &\le & \|y-y_k\| + \|y_k - \sum_j c_j(y_k)b_j \|+ \| \sum_j c_j(y-y_k)b_j \| \\ &=& \|y-y_k\| + 0+ \| \sum_j c_j(y-y_k)b_j \| \end{eqnarray} from which we get that $y=\sum_j c_j(y)b_j$ and so $y \in F$.

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