How can I think about a net in topology? [duplicate]

To begin a net over the directed set $\mathbb{N}$ is just a sequence. So nets are in a way a generalization of sequences. Let me motivate the analogy further.

Why do we need nets? Let $X$ and $Y$ be topological spaces. When $X$ is first countable the following conditions on a function $f: X \rightarrow Y$ are equivalent:

1. $f$ is continuous,
2. For every sequence $x_n \rightarrow x$ in $X$ we have $f(x_n) \rightarrow f(x)$.

When the first countability assumption is no longer true, the statement $2. \Rightarrow 1.$
is no longer valid in general. However the following equivalence holds in any case

1. $f$ is continuous,
2. For every net $x_\alpha \rightarrow x$ in $X$ we have $f(x_\alpha) \rightarrow f(x)$.

More generally, almost every property about the topology of metric spaces (where the space is first countable) which has an equivalent formulation using sequences, remains true in general topological spaces but one has to replace sequences by nets.

Let me explain why the property $\forall \alpha, \beta \in D: \exists \gamma \in D: \alpha, \beta \leq \gamma$ is crucial for nets to have any application. In topology it happens a lot that we have a sequence $(x_n)$ such that

1. $\exists N_1 \in \mathbb{N}: \forall n \geq N_1: x_n$ satisfies some property A,
2. $\exists N_2 \in \mathbb{N}: \forall n \geq N_2: x_n$ satisfies some property B.

If we choose $N = \max \{N_1,N_2\}$ then for all $n \geq N: x_n$ satisfies both property $A$ and $B$. This strategy is applied very often when $A$ and $B$ together with the triangle inequality give something meaningful.

The same strategy can be applied for nets precisely because of the property that I mentioned.