Show that in any topological space, a sequence converges if and only if it is eventually constant.

Question

I have some trouble in proving this statement.

In any topological space, a sequence converges if and only if it is eventually constant.

I struggle to show the direction to the right.

$(\Rightarrow)$ In a topological space $(X,\tau)$, a sequence $(x_n)\in X$ converges to $x\in X$ if for any open set $U$ there exists an $N$ such that $n>N$ implies $x_n\in U$. We want to show there exists $x\in X$ and $N$ such that $n>N$ implies $x_n=x$.

Suppose $(x_n)$ converges to $x$ and let the $x$ we about to show be the limit in the convergent sequence. Then how are we going to find the $N$? If we use the $N$ in the condition convergent sequence then we cannot guarantee $x_n=x$ for all $n>N$. Also should we pick a specific open set in the first place?

Could someone please give some help? Thanks.

Answer 1

"In any topological space, a sequence converges if and only if it is eventually constant."

Ah, this is ambiguous.

If it means: A sequence will converge in any topological space (in which it exists) if and only if it is eventually constant: then it is true.

If a sequence converges in any topological space it is defined, it will converge in a discreet topology. In a discreet topology all sets are open so $\{x_n\}$ is open. There is an $x$ so that for any open $U$ so that $x \in U$ implies there is an $N$ so that $n > N$ implies $x_n \in U$, then $\{x\}$ is such an open set so $n > N$ implies $x_n \in \{x\}$. So $\{x_n\}$ is eventually constant.

If a sequence is eventually constant then there is an $N$ so that $n > N$ implies $x_n = x$. For any open $U$ so that $x \in U$ then $x_n = x \in U$. So $\{x_n\} \rightarrow x$.

THe only way this could fail is if there is not open set $U$ that $x \in U$. But as the universal set is open and $x$ is in the universal set, this can not be.

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"I have some trouble in proving this statement. "

As well you ought. It's not true.

"We want to show there exists $x∈X $and $N$ such that $n>N$ implies $x_n=x$".

You can't. It's not true. $\{\frac 1n\} \rightarrow 0$ but $1/n = 1/m \iff n =m$. There is no $N$ so that $n, m > M \implies 1/n = 1/m$.

Answer 2

I suspect the statement you are trying to prove is $$\mbox{"If $A=\{a_n\}$ is a sequence from $X$, and $A$ converges in all topologies on $X$, then $A$ is eventually constant."}$$ This is a very different statement from what you've written! As fleablood wrote, there are lots of instances of topological spaces which have convergent, non-eventually-constant sequences. Do you understand, for example, why the sequence $1, {1\over 2}, {1\over 3}, {1\over 4}, ..., {1\over n}, . . .$ is convergent in the usual topology on $\mathbb{R}$?

So let's try to prove the correct statement above. It's easy to show that every eventually constant sequence is convergent (I think you've already done this?). In the other direction, suppose $A=\{a_n\}$ is a sequence in $X$ which is not eventually constant; we want to build a topology on $X$ such that $A$ is not convergent in that topology. Now think about the definition of convergence; intuitively, topologies with "more" open sets have "fewer" convergent sequences (because there are more requirements to satisfy for a sequence to be convergent - "for every open set containing $x$ . . ."). So: what is an example of a topology you know how to put on a set $X$, which has "lots" of open sets?