# A net converges to a point iff every subsequence of the net converges to the same point in first countable topological spaces

I'm having trouble proving the fact that in first countable topological spaces, a net converges to a point iff every subsequence of the net converges to the same point.

I first encountered this problem when I was learning dominated convergence theorem. I often tried to pass the limit under the integral sign, especially when dealing with partial derivative of an integral and using Leibniz's rule. The statement of DCT involves a sequence of functions and the convergence of a sequence of integrals. $$lim_{n \rightarrow \infty} \int f_n = \int lim_{n \rightarrow \infty} f_n$$But when I was dealing with partial derivative of an integral, it's a net of functions which gives the convergence of a net of integrals. For example, $$lim_{h\rightarrow0} \int_0^t \frac{H(t+h,s)-H(t,s)}{h}ds = \int_0^t lim_{h\rightarrow 0} \frac{H(t+h,s)-H(t,s)}{h}ds = \int_0^t \frac{\partial H(t,s)}{\partial t} ds$$

This is often proved by DCT.

I don't know why exactly we can do this. I was told it is because of the result that a net converges to a point iff every subsequence of the net converges to the same point in first countable topological spaces. I tried to prove this result by myself but couldn't get anywhere.

Thanks in advance for any help!

What is true, for instance, is that a net indexed by $(0,t)$ (with the reverse order) as in your example converges to a point $p$ iff every subsequence converges to $p$, in any space (no first countability required!). This is easy to prove if you just write down the definitions. The forward direction is trivial. For the reverse direction, if a net $(x_h)$ indexed by $(0,t)$ fails to converge to $p$, that means there is some open neighborhood $U$ such that for all $\epsilon>0$ there exists $h<\epsilon$ such that $x_h\not\in U$. Now for each $n$, choose some $h_n<1/2^n$ such that $x_{h_n}\not\in U$. Then $(x_{h_n})$ is a subsequence of $(x_h)$ which does not converge to $p$.