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Let $f\colon E\to \mathbb{R}$ be continuous at $p \in E$. Prove that there exists a positive constant $M$ and $\delta > 0$ such that $|f(x)| \le M$ for all $x \in E \cap N_\delta(p)$.

The book says to take $\epsilon = 1$, then there exists a $\delta > 0$ such that $|f(x) - f(p)| \le 1$, for all $x \in E \cap N_\delta(p)$.

How would I start this?

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  • $\begingroup$ What's $N_{\delta}$ ? $\endgroup$ Nov 19, 2013 at 19:13
  • $\begingroup$ Why do you have to take the intersection $E \cap N_{\delta}$ couldn't you just take $N_{\delta}$? I.e. is $E$ a subspace of some space? $\endgroup$ Nov 19, 2013 at 19:15

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Hint: Let $J$ be the open interval $J=(f(p)-1,f(p)+1)$. Since $f$ is continuous, you know that $f^{-1}\big(J\big)$ will be an open subset which contains $p$. In particular, there is some $\delta > 0$ such that if $N_{\delta}$ is the open ball around $p$ of radius $\delta$, then $f(x) \in J$. In particular if $x \in N_{\delta}$ then $|f(x) - f(p)|<1$ which implies that $|f(x)|\leq 1 + |f(p)|$. Can you finish from here?

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  • $\begingroup$ Would I be right in thinking something along the sorts of $|f(x)| \le 1 + |f(p)| \le M$? Sorry, I'm not too great with these $\endgroup$
    – Brian
    Nov 20, 2013 at 2:28
  • $\begingroup$ In fact, just let $M = 1 + |f(p)|$. Then you can say, if you choose $\delta$ as I described in the solution, that if $x \in N_{\delta}$ then $f(x) \in J$ and thus $|f(x)-f(p)| < 1$. Using the reverse triangle inequality, you get $|f(x)| - |f(p)|< 1$ implying that $|f(x)|< 1+|f(p)| = M$. Then you're done! $\endgroup$
    – Tom
    Nov 20, 2013 at 2:33
  • $\begingroup$ Cool! Thanks a lot for the help! $\endgroup$
    – Brian
    Nov 20, 2013 at 2:44

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