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If $f:\mathbb{N}\rightarrow{\mathbb{R}}$ :

$f(x)=0 $ , if $x=1$ and $f(x)=\dfrac{1}{x-1}$ if $x>1$

(i) Proof $ f(\mathbb{N})$ is a compact set.

(ii) Is Weierstrass's theorem applicable? $f^{-1}:f(\mathbb{N})\rightarrow{\mathbb{N}}$

Thanks!

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    What have you tried? For $(i)$, use Heine-Borel. Regarding $(ii)$, Weierstrass had a lot of theorems, which one do you mean? – 2017-02-11
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    Applicable to what? $f$ or $f^{-1}$? – 2017-02-11
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    is for $f^{-1}$ – 2017-02-11

1 Answers 1

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See that the set $f(\Bbb N)$ is closed and bounded thus conclude that it is compact. To see boundedness of $f(\Bbb N)$ see that $f(x) \leq 1 $ forall $x \in \Bbb N$.

$f(\Bbb N) = \{0,1,1/2,1/3, \cdots\}$. It is easy to see that $f(\Bbb N)$ is a closed set in $\Bbb R$ since $0$ is the only limit point of the set and it is in the set.

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    How proof closed? – 2017-02-11
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    you're right thanks. – 2017-02-11
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    Thanks for accepting my answer – 2017-02-11
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    I have not understand your question of part 2 – 2017-02-11