# On eventually constant sequences

It is of course true that in a discrete space a sequence converges iff it's eventually constant. Is the converse true, i.e., if the only convergent sequences in a space are eventually constant, is the space necessarily discrete? I want to examine this statement for metric spaces but use of Hausdorff spaces is always welcome.

• What are your thoughts? Dec 2 '13 at 23:08
• @Don If in a Hausdorff space such as the one I described there is a point such that the respective singleton is not open, for every open neighbourhood of x there is another point in the same neighbourhood. I don't know how to make this better. I have some ideas for metric spaces but I'm still trying them out. Dec 2 '13 at 23:14

No, the converse does not hold in general. Let $X$ be any uncountable set, and fix a point $p\in X$. Let $$\tau=\{U\subseteq X:p\notin U\}\cup\{X\setminus C:C\text{ is a countable subset of }X\}\;;$$

then $\tau$ is a non-discrete Hausdorff topology on $X$ in which the only convergent sequences are the trivial ones. (In fact $\tau$ is a $T_5$ topology: the space is hereditarily normal.)

Added: If $X$ is metrizable, however, the converse does hold. Suppose that $x\in X$ is not an isolated point. Then for each $n\in\Bbb N$ there is a point $x_n\in B(x,2^{-n})\setminus\{x\}$, and the sequence $\langle x_n:n\in\Bbb N\rangle$ is clearly a non-trivial sequence converging to $x$.

• Dr. Scott, does this topology have a name that you know of? I have seen it come up a lot to give counterexamples.
– LASV
Dec 2 '13 at 23:18
• Thank you for the answer, but my question was mainly for metric spaces and Hausdorff spaces were to enter the game if it was possible to prove the result without reference to the metric. EDIT: You covered me in your addition, thanks a lot! Dec 2 '13 at 23:20
• @Steve: See the addition to my answer. Dec 2 '13 at 23:20
• I saw that, I had just found out myself, too, but thanks for your interest, your example was also very interesting! Dec 2 '13 at 23:21
• @Luis: I don’t know of any standard or even common name. One might call it the one-point Lindelöfization of an uncountable discrete space, by analogy with the one-point compactification. Dec 2 '13 at 23:21

No. Take a look at $\mathbb{R}_{cocountable}$.

• Nice example but not Hausdorff, so... Dec 2 '13 at 23:12
• Ah, I did not read the Hausdorff condition... Thanks @StevePap.
– LASV
Dec 2 '13 at 23:13
• At least it answers the first part of his question...
– LASV
Dec 2 '13 at 23:14