Convergence of a sequence for the three-point set

Question

In Munkres Topology section 17 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 $c$.

I have read the explanation for this but am still finding it difficult to see why this is true, could someone give some further detail on why this happens please?

Thank you.

Answer 1

In asking a related question (which can be found here https://math.stackexchange.com/a/2959100/226937 ) I also found the answer to this question.

For $X=\{a,b,c\}$ with the trivial topology of $\ \mathcal{T}=\{∅,X\}\ $ the limit of a sequence will be all elements in the set $X$ since the smallest open set in the topology (i.e. the set $X$) which contains the limit point also contains all other points in $X$. This would also be the same for any set of a finite number of elements equipped with the trivial topology.

Answer 2

Let $X$ be any set. Consider the topology $\mathcal{T} = \left\{ \emptyset , X \right \}$. Let $p \in X$ be a point and let $x_n = p$ for all $n \in \mathbb{N}$.

Now suppose that $q \in X$. We will prove that $x_n$ converges to $q$. To see this, let $U$ be an open set containing $q$. Since $U \neq \emptyset$, $U = X$. Let $N=1$. Then for all $n>N$, we have $a_n =p \in X =U$. Since $U$ were arbitrary, we have proved that $a_n$ converges to $q$.

Answer 3

The topology there in the book is in fact $\{a,b\}, \{b\}, \{b,c\}\}$. So every open set in the topology contains $b$, so the sequence $b,b,b,\ldots$ converges to all $3$ points by the definition given in the text: take $N=1$ always.