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The title is pretty almost self explanatory, but I will leave my precise question below.

Question. Consider the normed space $(\Bbb R^n,\|\cdot\|_\infty)$, where the infinite norm is the usual one. Show that the unit sphere defined by $$ S = \{\alpha \in \Bbb R^n \colon \|\alpha\|_\infty = 1\}$$ is closed and bounded.

My attempt. $S$ is obviously bounded (each element of $S$ satisfies the property $\|\alpha\|_\infty = 1$ which implies that each element of $S$ also satisfies the property $\|\alpha\|_\infty \leqslant 1).$

Now, I was trying to prove $S$ is closed (directly) but I am having some troubles. To see $S$ is closed, it suffices to prove that $\overline{S} \subset S.$ Let $\alpha \in \overline{S}$ be an arbitrary point. Then, there exists a sequence $(\alpha_k)_{k \in \Bbb N}$ s.t. $\alpha_k \rightarrow \alpha$ as $k \rightarrow \infty.$ By definition of a convergent sequence,

$$\forall \epsilon > 0, \exists N = N(\epsilon) \colon \forall n \in \Bbb N, n > N \Rightarrow \|\alpha_n-\alpha\|_\infty < \epsilon. $$ But $$\|\alpha\|_\infty = \|(\alpha-\alpha_n) + \alpha_n\|_\infty \leqslant \|\alpha-\alpha_k\|_\infty + \|\alpha_n\|_\infty < \epsilon +1. $$ From here I don't know how to proceed. Any help would be apreciated.

Anne Bauval
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    The final step is to let $\varepsilon \rightarrow 0^+$ in the last inequality, since it holds for each $\varepsilon > 0$ (and don't forget to change the strict inequality to non-strict one!). – Zerox Nov 15 '22 at 10:50
  • This answered your question Showing the the unit sphere is closed using sequences. Note that you should similarily bound $|\alpha|_\infty$ from below – Anne Bauval Nov 15 '22 at 11:11
  • @AnneBauval Well, every norm is bounded below by $0$... –  Nov 15 '22 at 11:14
  • Oh, I see what you mean now. I would have to find something like $|\alpha|_\infty > 1 - \epsilon$. Is this it? –  Nov 15 '22 at 11:19
  • Thank you for your help. I think I will stick with @SomeCallMeTim answer though, much more clean and quick –  Nov 15 '22 at 11:20
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    Of course, but the more elementary way was not obvious to you and your question was about it. Glad that the comments helped you to solve your issue. Also, you could shorten your title and text to "Prove that the unit sphere of $\Bbb R^n$ is closed" (since the unit sphere of $\Bbb C^n$ is nothing but the unit sphere of $\Bbb R^{2n}$). – Anne Bauval Nov 15 '22 at 11:25

1 Answers1

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The essential property you are missing is that the norm is continuous (basically by definition). This means that your proof gets even simpler: Take $(a_n)_n$ so $a_n\to a$, with $a_n\in S$. Then, as the norm is continuous and therefore commutes with limits, $1=\lim_{n\to\infty}||a_n||_\infty =||\lim_{n\to\infty}a_n||_\infty=||a||_\infty $.

Another way to see it is that since $||\cdot||_\infty:\mathbb{K}^n\to\mathbb{R}_{\geq 0}$ is continuous, the preimage of closed sets are closed. And $S$ is exactly the preimage of $\lbrace 1\rbrace$.

SomeCallMeTim
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  • Oh wow. This makes a lot more sense than working with the $\epsilon$ definition of convergence. Thanks for you help. –  Nov 15 '22 at 10:59
  • @SomeCallMeTim I would have proved closedness the same way as you did. But Zero's comment is a tighter answer to Fire's question. (S)he has to practise calculus before diving into topology. – Anne Bauval Nov 15 '22 at 11:05