Understanding proof of sequential continuity?

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?

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.

• 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. – J. Doe Mar 18 at 3:25

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).