# Prove the sequential criterion for continuous functions

The theorem is: A function $$f: A \to \mathbb{R}$$ is continuous at the point $$c \in A$$ iff $$\forall x_n \in A$$ that converges to $$c$$, the sequence $$f(x_n)$$ converges to $$f(c)$$.

I know this question has been asked before, but I have a specific question in one of the directions that is not directly referenced in the other answers. At least, it didn't appear to be.

I believe I have the forward direction:

Suppose $$f$$ is continuous at $$c$$. This tells me

$$\forall \epsilon > 0, \exists \delta > 0,$$ s.t. whenever $$|x - c| < \delta, |f(x) - f(c)| < \epsilon$$.

Consider $$x_n \to c$$ where $$x_n \in A$$. This tells me that

$$\forall \rho > 0, \exists N \in \mathbb{N}$$ s.t. whenever $$n \geq N, |x_n - c| < \rho$$.

I want to show

$$\forall \epsilon > 0, \exists \delta' > 0$$ s.t. whenever $$|x_n - c| < \delta', |f(x_n) - f(c)| < \epsilon$$

Choose $$\rho = \delta' = \delta$$. Since $$x_n \to c$$, I have that $$|x_n - c| < \rho$$ for any $$\rho$$ I choose, including $$\delta'$$ and $$\delta$$. By making this choice, it follows from continuity of $$f$$ that $$|f(x_n) - f(c)| < \epsilon$$. This is exactly what I wanted to show.

Backwards direction:

Suppose $$x_n \in A$$ and $$x_n \to c$$. Also suppose that $$f(x_n) \to f(c)$$. I have the following two items:

$$\forall \rho > 0, \exists N_1 \in \mathbb{N}$$ s.t. whenever $$n \geq N_1$$, $$|x_n - c| < \rho$$

$$\forall \sigma > 0 \exists N_2 \in \mathbb{N}$$ s.t. whenever $$n \geq N_2$$, $$|f(x_n) - f(c)| < \sigma$$

I want to show

$$\forall \epsilon > 0, \exists \delta > 0$$ s.t. whenever $$|x - c| < \delta, |f(x) - f(c)| < \epsilon$$

Now, what I essentially want to happen is that every point in $$A$$ has a sequence converging to it. If that was the case, then by the convergence of $$x_n$$ and $$f(x_n)$$, I can take $$\delta = rho$$ and then $$\epsilon = \sigma$$. The problem is, I am not convinced this has to happen. Why couldn't there be $$y \in A$$ where $$|y - c| < \delta$$, but no sequence converges to $$y$$? After all, $$A$$ is just a set, we don't know if it's closed. I feel like $$A$$ need not contain all of its limit points just by being a set. Am I on the right track? Or is my approach completely off and this fact is the downfall.

• It is true that, for any $a\in A$, there is some sequence $(a_n)_{n\in\Bbb N}$ of elements of $A$ such that $\lim_{n\to\infty}a_n=a$. Just take $a_n=a$ for each $n\in\Bbb N$. Jul 10, 2022 at 20:57
• @JoséCarlosSantos Oh right, it would be a constant sequence of $a,a,a,\dots$. I think I can finish the problem now! I'll try to answer my own question. Jul 10, 2022 at 21:40

As pointed out in the comments, I can take the constant sequence $$a_n = a$$ for any $$a \in A$$. This is a sequence in $$A$$ and it converges to $$a$$.
Now, let $$\rho = \delta$$ and $$\sigma = \epsilon$$. Then, I have
$$|a_n - a| < \delta$$ and $$|f(a_n) - f(a)| < \epsilon$$. Therefore, the result holds.