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Let $f:\mathbb{R}\to\mathbb{R}$ be defined as $ f(x) := \begin{cases} x, & \text{if}\ x\in \mathbb N,\\\\ 0, & \text{else,} \end{cases} $

and $T=\mathbb{N}\cup\{n+1/n:n\in\mathbb{N}\}$.

The function $f$ is continuous on $\mathbb{N}$ with respect to the usual metric on the reals, as any function is continuous as every point is an isolated point. And $f$ is discontinuous on $T$ as $|f(n)-f(n+1/n)|=n>\epsilon=1/2$ say. Am I right?

TZakrevskiy
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Myshkin
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  • Every point in $T$ is still isolated, don't see the difference in reasoning from $\Bbb N$. – TZakrevskiy Sep 19 '13 at 08:37
  • I'd rather say "The restriction of $f$ to $\mathbb N$ is continuous" instead of saying "$f$ is continuous on $\mathbb N$". The latter wording implies that each point of $\mathbb N$ is a point of continuity of $f$ as a function from the reals to the reals which is certainly not true. – kahen Sep 19 '13 at 09:11
  • @kahen I am familiar with the wordings "cont. on $D$" where $D$ is a domain and $\left.f\right|_D$ is (uniformly) cont. and "cont. in $A$" meaning every point in $A$ is a point of continuity of $f$ on it's domain. – AlexR Sep 19 '13 at 09:24

1 Answers1

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Actually, your proof is correct but not written down properly.
Let $\epsilon = \frac{1}{2}$ then we claim there is no $\delta > 0$ with $$|f(x) - f(y)| < \epsilon \qquad \forall\ |x-y| < \delta$$ Let $\delta_0$ be such a choice and chose $N := \lceil \frac{1}{\delta_0} \rceil + 1$, then $$\left|N - N+\frac{1}{N}\right| = \frac{1}{N} < \delta_0$$ But $$\left|f(N) - f\left(N+\frac{1}{N}\right)\right| = N > \epsilon$$ So $f$ is not uniformly continuous on $T$, but locally continuous.

$f$ is (uniformly) continuous on $\mathbb N$, locally cont. on $T$ and discontinuous on $\mathbb R$ in every $x\in\mathbb N$

AlexR
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