When is sequence limit equivalent function limit?

Question

Let $f: X \to Y$ be a function on $X$ into $Y$ where $X, Y$ are topological spaces, define the function limit of $f$ at $x_0 \in X$ as follows

$$\lim_{x \to x_0} f(x) = y_0$$

where $y_0 \in Y$ if for each open set $U_Y \subseteq Y$ containing $y_0$, there exists an open set $U_X$ containing $x_0$ such that $f(U_X) \subseteq U_y$ ($\lim_{x \to x_0} f(x) = f(x_0)$ recovers the usual definition of continuous function)

Let $(x_k)_{k \in \mathbb{N}}$ be a sequence, define the sequential limit of $(x_k)_{k \in \mathbb{N}}$ as follows

$$\lim_{k \to \infty} x_k = x_0$$

where $x_0 \in X$ if for each open set $U_X \subseteq X$ containing $x_0$, there exists a $K_1 \in \mathbb{N}$ such that $x_k \in U_X$ for all $k \geq K_1$

Define the next limit

$$\lim_{k \to \infty} f(x_k) = y_0$$

where $y_0 \in Y$ if for each open set $U_Y \subseteq Y$ containing $y_0$, there exists a $K_2 \in \mathbb{N}$ such that $f(x_k) \in U_Y$ for all $k \geq K_2$

I am wondering when is $\lim_{k \to \infty} f(x_k) = y_0$ for all possible sequences $(x_k)_{k \in \mathbb{N}}$ enough to conclude that $\lim_{x \to x_0} f(x) = y_0$

@Ningxin please help me understand. You answered the question. Why did you leave the answer as a comment? This seems to be an epidemic on SE right now. — IAAW, Oct 20 '23 at 18:52
@IAAW I didn't answer the question. I didn't explain what "first countable" means, nor why it works. I gave a hint, a term that OP can search and find the actual answer himself.
Another reason is that it is not encouraged here to answer a question that has already been asked and answered many times. — Ningxin, Oct 20 '23 at 19:45
You can have a look at the Wikipedia article on sequential spaces. There are also various posts on this site related to the topic, such as: Continuity and image of convergent sequences Some post with a proof about first-countable spaces: Proofs about continuity and convergence in topological spaces, Proof for sequentially continuous function with a first countable domain is continuous and ... — Martin Sleziak, Oct 21 '23 at 16:26
... Theorem 30.1 (b) in Munkres' TOPOLOGY, 2nd ed: The sequential criterion for continuity. Some posts with countexamples: Give an example of a function $f :X \to Y$ which is sequential continuous but not continuous where $X$ and $Y$ are some topological spaces., Example of topological spaces where sequential continuity does not imply continuity and Example where $f$ is discontinuous. — Martin Sleziak, Oct 21 '23 at 16:26

score 0 · Accepted Answer · answered Oct 21 '23 at 07:29

If $X$ has a countable local basis $\mathcal{B}$ at $x_0$, i.e. there is a countable collection of open neighbourhood $\mathcal{B} = \{ B_k: k \in \mathbb{N} \}$ of $x_0$ such that for each open neighbourhood $U_X \subseteq X$ of $x_0$, there exists a $B_k \in \mathcal{B}$ such that $B_k \subseteq U$. We construct an open neighbourhood $A_n = \bigcap_{k=1}^n B_k$ of $x_0$ for each $n \in \mathbb{N}$. Then $A_1 \supseteq A_2 \supseteq A_3 \supseteq ...$ and $\mathcal{A} = \{A_k: k \in \mathbb{N} \}$ is also a countable local basis at $x_0$ as $A_k \subseteq B_k$.

Suppose the opposite, there exists an open neighbourhood $U_Y$ of $y_0$ such that for all open set $U_X$ containing $x_0$ then $f(U_X) \setminus U_Y \neq \emptyset$.

For each $A_k, k \in \mathbb{N}$, choose $\bar{x}_k \in A_k$ such that $f(\bar{x}_k) \notin U_Y$. The sequence $(\bar{x}_k)_{k \in \mathbb{N}}$ converges to $x_0$ as each open neighbourhood $U_X \subseteq X$ of $x_0$ contains an $A_k$ and that $A_k$ contains all $A_{k+1}, A_{k+2}, ...$.

However, $(f(\bar{x}_k))_{k \in \mathbb{N}}$ does not converges to $y_0$ since there exists an open neighbourhood $U_Y$ of $y_0$ such that $f(\bar{x}_k) \notin U_Y$ for all $k \in \mathbb{N}$. Contradiction.

Thanks to Ningxin for their hint.

When is sequence limit equivalent function limit?

1 Answers1