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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function which has the following property:

(1)$\quad$ Every sequence $x_n\rightarrow x$ has a subsequence $(x_{n_k})$ such that $f(x_{n_k})\rightarrow f(x)$.

Prove that

$$ (2)\quad \lim_{y\rightarrow x}f(y)=f(x) $$

Proof by contradiction. Suppose that (2) does not hold. Then, there exists $\varepsilon>0$ and a sequence $x_n\rightarrow x$ such that $|f(x_n)-f(x)|\geq \varepsilon$. This contradicts to the assumption (1).

My questions.

1) Could we give a direct proof for the above fact without using proof by contradiction?

2) Can we give a visual explanation (intuition) for the above fact?

Thank you for all the answers.

Blind
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    Think about the counterpositive, that is: If the limit $\lim_{y\rightarrow x}f(y)$ does not go to $f(x)$, you can define some sequence $(x_{n})$ such that $x_{n}$ converges to $x$ but $f(x_{n})$ does not converges to $f(x).$ Can you see that? – Mateus Rocha Dec 13 '19 at 00:29

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Well sure we could give a direct proof, take $\{x_n\},\{y_n\}$ sequences converging to $x$, such that $$f(x_n)\to \lim_{r\searrow0}\sup_{z\in(x-r,x+r)}f(z),$$ $$f(y_n)\to\lim_{r\searrow 0}\inf_{z\in(x-r,x+r)}f(z).$$ Then choose $\{{x_{n}}_k\},\{{y_n}_k\}$ subsequence with $$f({x_{n}}_k),f({y_n}_k)\to f(x).$$ Then since subsequences of convergent sequences converge to the same limit $$\lim_{r\searrow 0}\inf_{z\in(x-r,x+r)}f(z)=f(x)=\lim_{r\searrow0}\sup_{z\in(x-r,x+r)}f(z),$$ hence $\lim_{z\to x}f(z)$ exists and equals $f(x).$

Melody
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  • Why can we choose ${x_{n_k}}$ and ${y_{n_k}}$? – Blind Dec 13 '19 at 00:36
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    We're assuming if $z_n\to x$ for any sequence ${z_n},$ then we can choose a subsequence ${z_n}_k$ such that $f({z_n}_k)\to f(x).$ – Melody Dec 13 '19 at 00:37
  • Please explain why we can choose $x_n, y_n$? – Blind Dec 13 '19 at 00:46
  • @egorovik Take $$f:\mathbb{R}\to\mathbb{R}$$ to be $1$ on all irrationals, and $0$ on all rationals. Then $\lim_r\sup_{z\in(x-r,x+r)}f(z)=1\not=0=f(0).$ We cannot assume the limit inferior and limit superior are just $f(x),$ as that condition is equivalent to the limit existing, and being $f(x).$ That's circular logic. – Melody Dec 13 '19 at 00:46
  • Please explain clearer... – Blind Dec 13 '19 at 00:51
  • In my point of view, you use limsup and liminf to solve my problem? – Blind Dec 13 '19 at 00:55
  • @Blind It is a general property that you always have sequence converging to the liminf/limsup, and $\lim_{z\to x}f(z)=f(x)$ is equivalent to $$\lim_{r\to x}\sup_{z\in(x-r,x+r)} f(z)=\lim_{r\to x}\inf_{z\in(x-r,x+r)}f(z)=f(x).$$ – Melody Dec 13 '19 at 01:00
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    @Blind Here is a proof for the sequence limit case, we are dealing with a continuous limit here, but the proof there isn't too hard to modify for the continuous limit case.

    https://math.stackexchange.com/questions/581128/prove-that-subsequence-converges-to-limsup

    Here's one about the equivalence of the limits

    https://math.stackexchange.com/questions/122755/sequence-converges-iff-limsup-liminf

    I don't have space here for a detailed reply, but hopefully the links help.

     – Melody Dec 13 '19 at 01:04