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.

  • $\begingroup$ 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 $n\in\Bbb N$. $\endgroup$ Jul 10, 2022 at 20:57
  • $\begingroup$ @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. $\endgroup$
    – Nolan P
    Jul 10, 2022 at 21:40

1 Answer 1


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.


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