# Let $(x_\alpha)_{\alpha\in J}$ be a net in a topological space $X$. Show that if $J$ is a ﬁnite set then the net converges? …

Let $(x_\alpha)_{\alpha\in J}$ be a net in a topological space $X$. Show that if $J$ is a ﬁnite set then the net converges? ...

Why does it have to converge?

I don't understand why with nets this is true, whilst with sequences it isn't. if the value alternates between two values shouldn't it never converge?

• You are confused between the set J being finite and the values the sequence can take being a finite set. The sequence $\{1,1,1...\}$ is an infinite sequence even though every element in the sequence is just 1. – R_D Apr 18 '16 at 6:58
• so for example: X=Z+ J= x part of Z between (1-10) and the sequence be : if j is even, 1 if j is odd 0 is this not a net? – leo d torchia Apr 18 '16 at 7:19
• If $J$ is finite then, because it's directed, it has a greatest element $j_m$, and then the net converges to $x_{j_m}$. – BrianO Apr 18 '16 at 10:04

By the definition of the concept of "net", $J$ must be a directed set. Being finite, it must have a unique maximal element $j_\infty$, i.e. $j \le j_\infty \ \forall j \in J$ (intuitively, you can "reach" infinity, which you cannot do in $\Bbb N$). Then $x_{j_\infty}$ is the limit of the net.
To clarify the question in the comment: let $J = \{1, 2, \dots, 10\}$ with the usual order. Let $x_1 = x_3 = \dots = x_9 = 0$ and $x_2 = x_4 = \dots = x_{10} = 1$. Let us show that $0$ cannot be the limit of this net.
Take a neighbourhood $V$ of $0$, small enough such that $1 \notin V$. The definition of "limit of a net" says that there should exist $n_V \in J$ such that for every $n \ge n_V$ we should have $x_n \in V$. Notice that no matter who $n_V$ would be, clearly $10 \ge n_V$. But do we have $1 = x_{10} \in V$? Of course not, by how we chose $V$. This shows that $0$ does not satisfy the definition of "limit of a net".
On the other hand, $1$ does satisfy the definition: choose any neighbourhood $V$ of $1$ and always choose $n_V = 10$. Then all the terms with index greater than $10$ (which in fact means just $x_{10}$) are in $V$, and you just don't care about the first $9$ terms.