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.

  • $\begingroup$ What the heck does "eventually constant" mean? So since {1/n} converges to 0, I take it that means {1/n} is "eventually constant". What doesn't that mean? Any meaningful definition I can think of is either false or equivalent to the definition of convergence. $\endgroup$
    – fleablood
    Sep 25, 2016 at 23:50
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    $\begingroup$ Do you mean "any discrete topological space"? Because otherwise the statement you are trying to prove is false. (There's also the possibility that you're being asked to show that "a sequence converges in any topological space iff it is eventually constant", which is a subtly different statement - it's saying "the sequences which converge no matter the topology are the eventually constant ones".) $\endgroup$ Sep 25, 2016 at 23:51
  • $\begingroup$ @NoahSchweber I mean any topological space, not neccesarily discrete. Why is it false? $\endgroup$
    – user338393
    Sep 25, 2016 at 23:54
  • $\begingroup$ @NoahSchweber You're probably right. But the statement you said is it an iff statement? $\endgroup$
    – user338393
    Sep 25, 2016 at 23:57
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    $\begingroup$ It's false because it is not true. Why would it be true? Obviously there are sequence that converge that are not constant. ({1/n} for instance. The finite decimals expanding to pi for example.) That 's the entire bases of real analysis. Reals have the least upper bound property. There are bounded infinite sets with no max. So it's always possible to make converge sequence that are not constant. $\endgroup$
    – fleablood
    Sep 25, 2016 at 23:58

2 Answers 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?

  • $\begingroup$ The discrete topology has "lots" of open sets? But the statement asks for all topologies, not just discrete right? $\endgroup$
    – user338393
    Sep 26, 2016 at 0:12
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    $\begingroup$ @user338393 We're trying to prove the direction "If a sequence converges in every topology, then it is eventually constant." We're going to prove the contrapositive: "if a sequence is not eventually constant, then there is some topology in which it is not convergent." For this, we start with a non-eventually-constant sequence and try to construct one topology in which it does not converge. (And yes, the discrete topology is the right answer. Do you see how to show that a non-eventually-constant sequence doesn't converge in the discrete topology?) $\endgroup$ Sep 26, 2016 at 0:32

"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.


"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 $n, m > M \implies 1/n = 1/m$.

  • $\begingroup$ Thanks. There is a part that I don't quite get which is why "If a sequence converges in any topological space it is defined, it will converge in a discrete topology."? $\endgroup$
    – user338393
    Sep 26, 2016 at 0:40

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