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It suffices to show that any proper subspace M of X is closed, since if M is not proper the result is trivial.

I am unsure how to approach this proof. Contradiction seems a little messy, as supposing M is not closed implies M is i.) open or ii.) neither closed nor open. Any help would be much appreciated.

mattnz
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    It would indeed "suffice to show that any proper subspace M of X is closed", but this is false if X is infinite dimensional. – Anne Bauval Feb 19 '24 at 09:12

2 Answers2

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Hint: The easiest approach here is sequence continuity. That is, select an arbitrary Cauchy sequence $\{x_n\}_{n=1}^\infty \subset M$ and show that its limit lies in $M$.

Now, what does "finite dimensional" mean, and what does that mean for our sequence $\{x_n\}$?

Ben Grossmann
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  • Excuse my subscripts - I'm not sure how to do these properly! Choose some basis e_1,...,e_k for M. Then for each n, x_n is some linear combination of these basis vectors. So express (x_n - x_m) as a linear combination of these vectors. Then the coefficients of (x_n - x_m) form Cauchy sequences in R (or C) and thus converge, say, to a_1,...,a_k (as R, C are complete). Now take x = a_1e_1 + ... + a_ke_k. Then |x_k - x| tends to zero as k tends to infinity, so {x_n} converges to x (which is in M). Thus M is complete, and hence closed. Something along these lines? – mattnz Sep 10 '14 at 06:27
  • That's about the gist of it. Some help with the subscripts: I'd go with either $$ x_n = \sum_{i=1}^k a_i^{(n)} e_i $$ or the slightly less unwieldy $$ x(n) = \sum_{i=1}^k a_i(n) e_i $$ Note that you can go from Cauchy coefficients to a Cauchy sequence using the triangle inequality, but that this step fails for infinitely many coefficients. – Ben Grossmann Sep 10 '14 at 13:07
  • Right! So that's the reason this fails for an infinite dimensional subspace then? – mattnz Sep 10 '14 at 23:50
  • @mattnz that's right. If you want a concrete example where this fails, consider the space of sequences ${x_i}_{i=1}^\infty$ for which $x_i = 0$ for all but finitely many $i$. For a non-convergent Cauchy sequence, take $$ x_i(n) = \begin{cases} \frac 1i & i \leq n\ 0 & i > n \end{cases} $$ – Ben Grossmann Sep 11 '14 at 01:24
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Let $\dim M = n$. Then, identifying $M$ with $\mathbf{R}^n$, the norm of $X$ restricted to $M$ is equivalent to any of the standard norms on $\mathbf{R}^{n}$. (By equivalent norms I mean ones such that each is bounded by a constant multiple of the other.) Thus $M$ is a complete metric space.

It follows from this that $M$ is a closed subset of $X$. (For let $a_n$ be any sequence of elements in $M$ that converges to $a \in X$. It is a Cauchy sequence, and therefore has a limit in $M$. Thus $a \in M$.)

Dave
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