Reading the book about topology, I find an interesting difference between two spaces:

We use net convergence $\{p_\lambda\}_{\lambda\in\Lambda}\rightarrow p$ in frontier, but use sequence convergence $\{p_n\}_{n\in\mathbb N}\rightarrow p$ in latter.

So I want to clarify the cause of difference. My thought is that may be the first countable property. In Lee's book, there are two problems:

  • Any metric space is first coutable.

  • Let $X$ be first countable. We will have two statements:

    • For any set $A\subset X$ and any point $p\in X$, $p\in\bar A$ if and only if there is a sequence $\{p_n\}_{n\in\mathbb N}\in A$ such that $p_n\rightarrow p$.
    • A map $f:X\rightarrow Y$ is continuous if and only if $f$ takes convergent sequence in $X$ to convergent sequence in $Y$.

So we can substitute sequence convergence for net convergence in first countable space. However I do not think two problems give the core answer to my question. Any help, thank you.

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    $\begingroup$ I think that you mistakenly used the word "frontier" where you actually mean "former". (Good question anyway) $\endgroup$ – Giuseppe Negro Apr 20 '14 at 8:36
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    $\begingroup$ Relevant threads: this, and possibly that. The former might even be a duplicate. $\endgroup$ – Asaf Karagila Apr 20 '14 at 8:41

There are actually a couple of weakenings to first-countability for which it could be said that "sequences suffice".

  • A topological space $X$ is called Fréchet-Urysohn if for each $A \subseteq X$ and every $x \in \overline{A}$ there is a sequence in $A$ which converges to $x$.
  • A topological space $X$ is called sequential if a subset $A \subseteq X$ is closed iff it contains the limits of all convergent sequences of points in $A$.

It is not too difficult to show that first-countable $\Rightarrow$ Fréchet-Urysohn $\Rightarrow$ sequential (and neither arrow reverses).

Moreover, we have the following:

Fact. Suppose $X$ is a sequential space, and $Y$ is an arbitrary topological space. Then a function $f : X \to Y$ is continuous iff for every convergent sequence $\langle x_n \rangle_{n \in \mathbb{N}}$ in $X$ and every limit $x$ of this sequence, we have that $f(x)$ is a limit of the sequence $\langle f(x_n) \rangle_{n \in \mathbb{N}}$.

So in the larger class of Fréchet-Urysohn spaces we get the two results you mentions in the OP. (I say "every limit" and "is a limit" in the above because I am not restricting myself to Hausdorff spaces, where every convergent sequence/net has a unique limit.)

The basic idea of using net convergence in general topological spaces is to abstract away the family of (open) neighbourhoods of a point. More particularly, if given any point $x$ in a topological space $X$, he have that $$\mathcal{N}_x = \{ U \subseteq X : U\text{ is an open neighbourhood of }x \}$$ is directed by $\supseteq$. Moreover, if $A \subseteq X$, then $x \in \overline{A}$ iff $U \cap X \neq \varnothing$ for every $U \in \mathcal{N}_x$. So if you pick some $x_U \in U \cap A$ for each $U \in \mathcal{N}_x$, we get a net in $A$ converging to $x$. The next step would be to see that if you instead choose a neighbourhood base $\mathcal{B}_x$ at $x$, we still get a set which is directed by $\supseteq$, and the result still holds.

For first-countable spaces, we get the added bonus that every point has a countable descending basis $U_0 \supseteq U_1 \supseteq \cdots$, which in terms of the $\supseteq$-order is isomorphic to $\mathbb{N}$ with the usual order. This is why sequences suffice.

For Fréchet-Urysohn spaces which are not first-countable, it is not possible to work with fixed countable descending neighbourhood bases, but some other argumentation is needed. For example, in the space I describe here, it is the connection between the space and the real line (with the usual topology) that allows us to see that it is Fréchet-Urysohn.


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