Continuity in a normed space [closed]

Let $X$ be a normed space. Show that the function $f:X \to R$ defined by $f(x)=\|x\|$ is continuous on $X$.

-

closed as off-topic by Jonas Meyer, Ittay Weiss, Grigory M, Najib Idrissi, Martin SleziakJan 18 at 18:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Ittay Weiss, Grigory M, Najib Idrissi
If this question can be reworded to fit the rules in the help center, please edit the question.

I deleted my answer, just in case you're sitting an exam right now. Will undelete it in an hour. –  Rudy the Reindeer Oct 16 '12 at 16:22
Metric is Continuous - ProofWiki; Is a norm a continuous function? - MSE –  Martin Sleziak Oct 16 '12 at 16:24
no no I'm not doing an exam now...I'm trying to do some problems in continuity. –  ccc Oct 16 '12 at 16:25
Well, seeing something like "can somebody please help me with this problem immediately???" in a question is certainly unusual. I guess this is the reason behind Matt's reaction. –  Martin Sleziak Oct 16 '12 at 17:13
Welcome to MSE! Generally, questions that give background, motivation, and context receive more attention and upvotes. When the asker shows their work, readers are generally more inclined to respond with more and fuller answers. For more information, I recommend reading the faq. –  mixedmath Oct 16 '12 at 17:16

Suppose that $\{x_n\}\subset X$ and $x_n\rightarrow x$. Then $\|x_n-x\|\rightarrow 0$. We have $$|f(x_n)-f(x)|=|\|x_n\|-\|x\||\leq \|x_n-x\|.$$ Hence $f(x_n)\rightarrow f(x)$. Therefore $f$ is continuous.
I guess you want to use the $\varepsilon$-$\delta$-definition of continuity at $x_0$. Let $\varepsilon > 0$. We want to find $\delta > 0$ such that $$\|x_0 - x \| < \delta \implies |f(x) - f(x_0)| < \varepsilon$$
Here we have $|f(x) - f(x_0)| = | \|x\| - \|x_0\| |$. By the inverse triangle inequality we have $| \|x\| - \|x_0\| | \leq \|x - x_0\|$. Hence for $\delta = \varepsilon$ the claim follows.
No problem : ) ${}{}$ –  Rudy the Reindeer Oct 16 '12 at 16:21