Any complete subset of normed vector space is closed

Consider a normed vector space $(V,\lVert \cdot \rVert)$. Need to show that if $S\subseteq V$ is complete then $S$ is closed.

My approach:

A complete subset $S$ of $V$ satisfies that any sequence contained entirely in $S$ converges to a point in $S$, with respect to $\lVert \cdot \rVert$. Suppose $V$ is open. Then there exists a point $x\in V\backslash S$ such that $x$ is a limit point of $S$. But this contradicts the above definition. Hence, $V$ must be closed.

Please let me know if this approach is correct.

• The correct assumption to begin a proof of the contrapositive would be that $S$ is non-closed, not that $S$ is open (I think you’ve mixed up $S$ and $V$). I also think you are doing yourself a disservice by proving the contrapositive. Closedness is more or less contained in the very definition of completeness. – Zach Blumenstein Oct 21 '16 at 0:51
• Your definition of complete is wrong. $S$ is complete if any Cauchy sequence has a limit in $S$. – Hayden Oct 21 '16 at 0:51
• @sequence The way you wrote it is "any sequence contained entirely in $S$ converges to a point in $S$". In other words, you said any sequence converges, which isn't true. To answer your questions, yes to the first, no to the second. – Hayden Oct 21 '16 at 1:02
• @sequence While not incorrect, you keep saying "if a sequence converges in $S$...". But you don't need this assumption. If it converges at all, it is Cauchy. This is an important distinction if you're trying to talk about points which are outside of $S$, as you are doing. – Hayden Oct 21 '16 at 1:08
• @sequence Yes, that sounds much better. As in my answer below, you can also proceed without doing it by contradiction at all, but the flavor of the proof doesn't change. – Hayden Oct 21 '16 at 1:12

Suppose $S\subset V$ is complete. If $x$ is a limit point of $S$, then there is a sequence $(x_n)_{n\in\mathbb{N}} \subset S$ converging to $x$. But $(x_n)_{n\in\mathbb{N}}$ is thus Cauchy, so by the completeness hypothesis, converges inside of $S$. In other words, $x\in S$, so $S$ is closed.
(Note that nothing is special about being in a normed space here; this works for any metric space $(V,d)$.)