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A function $f\colon X\to \mathbb{R}$ is continuous at $a\in X$ if and only if we have $\lim{f(x_n)}=f(a)$ for every sequence $\{x_n\}_{n\in\mathbb{N}}$ in $X$ with $\lim{x_n}=a$.

Scratch.

($\Rightarrow$). As $f$ is continuous at $a\in X$, then for all $\varepsilon>0$ we're going to have an interval $(f(a)-\varepsilon,f(a)+\varepsilon)$ such that $(f(a)-\varepsilon,f(a)+\varepsilon)\cap f[X]\neq \emptyset$, so $f(a)$ is accumulation point for some sequence $\{y_n\}$ in $\mathbb{R}$.

But how do I proceed from here to show that there's also a sequence $\{x_n\}$ in $X$ such that $\lim{x_n}=a$? In order to be a sequence there, we're going to allow say $\delta>0$ to be any value, but $\delta$ is dependent on $\varepsilon>0$, as the definition of continuity tells us.

And after I do this, how to show that $\lim{f(x_n)}=f(a)$? Because although $f(a)$ is an accumulation point, it needs to be an accumulation point for the image of $f$ with $\{x_n\}\subset X$

($\Leftarrow$). Again, I'm having a hard time connecting the fact that a sequence $\{x_n\}$ in $X$ would need an $\delta>0$ to be any $\delta$ such that there'll be $N\in\mathbb{N}$ with $n>N\implies \left|x_n-a\right|<\delta$. And this $\delta$ will define our $\varepsilon>0$...

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    $\begingroup$ You're are misreading the statement you're trying to prove. For $\implies$ you need to prove that " If $f$ is continuous at $a$ then we have $\lim_{n \to \infty} f(x_n) = f(a)$ for any sequence $x_n$ such that $\lim_{n \to \infty } x_n = a$". To prove the statement you must 1. assume that $f$ is continuous at $a$ 2. consider any sequence $x_n$ such that $\lim_{n \to \infty} x_n = a $ and 3. Deduce from 1 and 2 that $\lim_{n\to \infty} f(x_n) = f(x) $ There is no need to show that $x_n$ exists. $\endgroup$ – Digitallis Nov 18 '20 at 1:30
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This is a corollary of a more general result: $f(x)\to b$ when $x\to a$ iff $f(x_n)\to b$ for each $(x_n)$ such that $x_n\to a$.

($\Rightarrow$). Consider $(x_n)$ such that $x_n\to a$ and $\epsilon>0$ arbitrary.

  1. Since $f$ is continuous, there exist $\delta>0$ such that if $|x_n - a|<\delta$ then $|f(x_n) - b|<\epsilon$.
  2. Since $x_n\to a$, there exists $n_0$ such that if $n\ge n_0$, then $|x_n - a|<\delta$.

Therefore, there exists $n_0$ such that $n\ge n_0$ implies $|f(x_n)-b|<\epsilon$.

($\Leftarrow$). Suppose the $f(x)\not\to b$. This means there exists $\epsilon>0$ such that for any $\delta>0$ there exists $x_\delta$ such that $|x_\delta - a|<\delta$ and $|f(x_\delta) - b|\ge \epsilon$.

In particular, for each $n\in\mathbb N$ we may take $\delta_n = \frac 1n$ and consider the sequence $(x_n)$ given by $x_n = x_{\delta_n}$. In this case, $|x_n - a|<\frac 1n\to0$ but $|f(x_n) - b|\ge\epsilon$ and therefore $f(x_n)\not\to b$, which is a contradiction.

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