0

Let $(X,\|\cdot\|)$ be a normed space. Prove that every sequence in $S_X=\{x\in X\mid \|x\|=1\}$ converges in $S_X$.

My attempt.

Let $(x_n)\in S_X$. Then, $\|x_n\|=1$ for all $n$. Now assume that $x_n\to x$. Given $\varepsilon>0$, there is $N\in\mathbb N$ such that $\|x_n-x\|<\varepsilon$ whenever $n\geqslant N$. Thus, $$\|x\|=\|x-x_n+x_n\|=\|x_n-x\|+\|x_n\|<\varepsilon +\|x_n\|=\varepsilon+1.$$

As $\varepsilon$ is arbitrary, it follows that $\|x\|=1$.

If the reasoning is correct, this shows that $S_X$ is closed. There have been many questions about this, but they involve showing that its complement is not open and so on. So, can someone please verify that my reasoning above is correct? Should I perhaps argue by contradiction assuming that $x\notin X$ or is this fine?

Thanks in advance.

inZugzwang
  • 1,835
  • Closed means $convergent$ sequences in $X$ converge in $S_X$. There are nonconvergent sequences in $S_X$. You want to use the "reverse triangle inequality" here too. – mathematician Dec 27 '15 at 16:31
  • 1
    I think that at this point you can conclude that $||x||\leq 1$ but not $||x||=1.$ – Balloon Dec 27 '15 at 16:32
  • @mathematician Did you mean convergent sequences in $S_X$ converge in $S_X$? This is the definition I was given. I took a convergent sequence in $S_X$ and I want to show that its limit has norm 1. Apparently, what I did above is wrong. So, how can I show this? – inZugzwang Dec 27 '15 at 16:40
  • @Baloown So, what do you suggest? With the reverse triangle inequality I get $|x|>1-\varepsilon$ but I don't see how this helps. – inZugzwang Dec 27 '15 at 16:45
  • 1
    @Ryuky If for all $\epsilon>0$ we have $1-\epsilon<||x||<1+\epsilon$, then we must have that $||x|| = 1$. – Marc Dec 27 '15 at 16:48
  • @Ryuky : I suggest to listen Marc ! – Balloon Dec 27 '15 at 16:49

1 Answers1

2

A more general way to prove this is to derive the result that any norm is continuous, see for example this question. If you take this result as given, then $$ ||x|| = ||\lim_{n\rightarrow\infty}x_n|| = \lim_{n\rightarrow\infty}||x_n|| = \lim_{n\rightarrow\infty}1 = 1. $$

Community
  • 1
Marc
  • 6,856