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I'm trying to understand proof of the following statement:

Q. Let $f$ be a function on a closed bounded interval $[a,b]$. Prove that $f$ is continuous at $ c \in [a,b]$ if and only if $f(x_n) \to c$ for any sequence of points $x_n \in [a,b]$ converging to $c$.

Here's the part of the proof for the direction $f$ is continuous at $c \implies f(x_n) \to f(c)$:

Pf. Suppose $f$ is continious at $c$, and let $\left\{x_n\right\}$ be a sequence which converges to $c$. The first means that for every $\epsilon > 0$ there exists $\delta_{\epsilon}>0$ such that $|f(x)-f(c)|< \epsilon$ whenever $|x-c|< \delta_{\epsilon}$. On the other hand since $x_n \to c$, there exists $n_{\epsilon}$ such that $|x_n-c| < \delta_{\epsilon}$ for all $n \ge n_{\epsilon}$. Then $\color{blue} {|f(x_n)-f(c)|< \epsilon}$ whenever $n \ge n_{\epsilon}$. By definition of convergence this means that $f(x_n) \to f(c)$.

My question is: how did the two definitions lead to the blue inequality? We have $|x_n-c| < \delta_{\epsilon}$ and we have $|x-c| < \delta_{\epsilon} \implies |f(x)-f(c)|< \epsilon $. Are $|x_n-c| < \delta_{\epsilon}$ and we have $|x-c| < \delta_{\epsilon}$ saying the same thing? I'd appreciate if someone could explain this.

EDIT: Perhaps I'm thinking of this wrong, but what I was expecting is that from this:

$$\begin{cases} |x_n-c| < \delta_{\epsilon} \\ |x-c| < \delta_{\epsilon} \implies |f(x)-f(c)|< \epsilon \end{cases} \implies |f(x_n)-f(c)|< \epsilon. $$

The second implication there would fall out purely algebraically?

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Let's think about what the two statements are saying in simple terms:

The first means that for every $\epsilon > 0$ there exists $\delta_{\epsilon}>0$ such that $|f(x)-f(c)|< \epsilon$ whenever $|x-c|< \delta_{\epsilon}$.

This means that for every $\epsilon$, there is an $\delta_\epsilon$ such that whenever the second condition is satisfied, that is whenever we have points $x$ such that $|x - c| < \delta_\epsilon$ holds, then we have $|f(x) - f(c)| < \epsilon$. So this is a statement about what we can say about $f(x)$ with regards to $f(c)$ whenever points $x$ satisfies a particular property.

On the other hand since $x_n \to c$, there exists $n_{\epsilon}$ such that $|x_n-c| < \delta_{\epsilon}$ for all $n \ge n_{\epsilon}$.

What this second part is showing is that the property that we wants holds for certain $x_n$'s when $n \geq n_\epsilon$. Basically this is showing that the $x_n$'s have the property we desire, and thus the conclusion of the first statement holds.

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  • $\begingroup$ I understood that the property holds for $x_n$'s for $n \ge n_{\epsilon}$ but couldn't convince myself the reasoning isn't dodgy because it felt like replacing/substituting $x$ with $x_n$. Thanks for your clarification. $\endgroup$ – J. Doe Mar 18 at 3:25
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It comes from the definition of continuity. Since the $x_n$ converge to c, we can get the terms to be as close to c as we want. Once the terms get within $\delta_\epsilon$, then f($x_n$) is within $\epsilon$ of f(c).

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