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Let $f:\Bbb N \to\Bbb R$ by writing $f(n) = \frac1{n^2}$. Is $f$ continuous at any point in its domain.

So, my thought is, $f$ is a function with domain $\Bbb N$ - natural numbers, so each point in the domain of $f$ is isolated.

So, according to the Limit Version of continuity, the function $f$ is contiuous at $x_0$ provided that $x_0$ is isolated in $A$, or else, $x$ is a point of accumulation (and $\Bbb N$ has none) so

$$\lim_{x\to x_0}f(x) = f(x_0)\;.$$

So $f$ is continuous at any point in its domain.

I'm really not too sure about my answer (or the way I answered it). So please let me know what you think. Thanks

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If you are working with $\,\Bbb N\,$ with the topology inherited from the usual euclidean one in $\,\Bbb R\,$ ,then $\,\Bbb N\,$ is a discrete space, and any function from such a space to any other space is always continuous. –  DonAntonio Nov 1 '12 at 2:43
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2 Answers

up vote 1 down vote accepted

each point of $\mathbb{N}$ is isolated point, Hence continuos

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Let $\,X,Y\,$ be topological spaces, with $\,X\,$ a discrete space. Then, any function $\,f:X\to Y\,$ is continuous.

Hint: any subspace of $\,X\,$ is open...

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