Convergence of a constant sequence in a set of finite elements

Question

In Munkres Topology section 16 in the subsection on Hausdorff Spaces there is a motivating example involving the three-point set $\{a,b,c\}$ which states that the sequence defined by setting

$x_n=b$

converges to $a$, $b$ and $c$.

In any finite set of elements would a constant sequence always converge to all elements in the set?

If you could also provide any examples or counter examples that would be helpful.

Thank you.

Answer 1

A constant sequence $(a_n)$ always converges to the constant value $a=a_n$. That doesn't depend on the underlying topology.

Any other limit depends on topology. In particular if $U$ is the smallest open subset containing $a$ then every element of $U$ is a limit point of $(a_n)$. Such $U$ doesn't have to exist of course, but in finite spaces it always does (or in other words: finite spaces are Alexandrov spaces). Of course there might be more limit points then $U$.

For example, let $X=\{1,2,3\}$. We will consider different topologies on $X$:

1. $\mathcal{T}=\{\emptyset, X\}$, i.e. the anti-discrete topology. Then any point is a limit of any constant (or not) sequence;
2. $\mathcal{T}=\mathcal{P}(X)$, i.e. the discrete topology. Then any constant sequence has exactly one limit - the constant value. In fact for finite spaces this is equivalent: every convergent sequence has exactly one limit if and only if $X$ is discrete;
3. $\mathcal{T}=\{\emptyset, X, \{1,2\}\}$. Then each sequence $a_n=1$ and $a_n=2$ has three limit points: $1,2,3$, while the sequence $a_n=3$ has only one limit point, namely $3$.
4. $\mathcal{T}=\{\emptyset, X, \{1,2\}, \{3\}\}$. Then the sequence $a_n=3$ again has exactly one limit point: $3$. But now sequences $a_n=1$ and $a_n=2$ have two limit points each, namely $1,2$.

Each example can be directly calculated from the definition of a limit point and I encourage you to do that.

Answer 2

If the topology is Hausdorff, a convergent sequence converges to a unique limit. If for $x_n\rightarrow x$ you have an open neighbourhood $U$ such that $x\in U$ and $y\notin U$, then $x_n$ does not converge to $U$.

Furthermore a topology is Hausdorff if and only if limits of nets (generalization of sequences) are unique.

Answer 3

A simple observation: If $X$ is a finite space, TFAE:

1. $X$ is discrete.
2. $X$ is Hausdorff.
3. Every constant sequence has a unique limit.
4. $X$ is $T_1$.

Proof

$1 \to 2$: Is clear as all discrete spaces are trivially Hausdorff.

$2 \to 3$: Is a standard result for all sequences (not just the constant ones)

$3 \to 4$: If there'd be an $x \in X$ such that $\{x\}$ is not closed, then there'd be some $y \in X, y \neq x$ with $y \in \overline{\{x\}}$, which means that every neighbourhood of $y$ contains $x$ as well. This implies that the constant sequence with value $x$ converges to both $x$ and $y$, contradicting $3$. So all singletons are closed and $X$ is $T_1$.

$4 \to 1$: As $X$ is finite (here we use it), all subsets are closed (and thus also open) when all singletons are.

In short, any non-discrete finite space will have a constant sequence converging to another point besides its own constant.