converging sequences in new topology

Question

Suppose $(X,T_X)$ is a topological space and $\infty_X \notin X$. Write $X^* = X \cup \{\infty_X\}$ and suppose the open sets of $X^*$ the empty set and the union of an open set in $X$ and the point $\{\infty_X\}$.

The statement I want to prove is that a sequence $(x_n)_{n \in \mathbb{N}}$ converges in our new space $(X^*, T^*))$ if and only if exactly one of the following conditions is fulfilled:

$\begin{align} &1) (x_n)_{n \in \mathbb{N}} \text{ becomes the constant sequence } \{\infty_X\} \text{ after a finite amount of terms } \\ &2) \text{ the subset of } (x_n)_{n \in \mathbb{N}} \text{ after leaving out all the points } \infty_X \text{ converges in } (X,T_X) \end{align}$

I know that the definition of a sequence converging to $x$ is that for every open neighborhood $U$ of $x$ there exists a $M \in \mathbb{N}$ such that $x_n \in U$ for any $n \geq M$. I'm struggling to see why this is equivalent to one of the two conditions.

Any help is appreciated!

Answer 1

The sequence $(x_n)_n$ converges if for every open set $O$ the sequence eventually stays in $O$, i.e. there is some $N \in \mathbb{N}$ such that for all $n > N$ we have $x_n \in O$.

Clearly, if $(x_n)_n$ is eventually constantly $\infty_X$ it converges to $\infty_X$. So the tricky part is to show the equivalence of "$(x_n)_n$ converges in $X^*$" and "$(x_{n_k})_k$ converges in $X$" for sequences $(x_n)_n$ that are not eventually constantly $\infty_X$. Here, I denote $(x_{n_k})_k$ the sequence that we get from $(x_n)_n$ by removing all $\infty_X$'s.

So assume $(x_n)_n$ is as described above and it converges in $X^*$. Let $U \subset X$ be open in $X$. Then $U \cup \{\infty_X\}$ is open in $X^*$ and therefore $(x_n)_n$ eventually stays in $U \cup \{\infty_X\}$. But then it follows immediately that $(x_{n_k})_k$ eventually stays in $U$. Thus, $(x_{n_k})_k$ converges in $X$.

In the other direction, assume that $(x_{n_k})_k$ converges in $X^*$. Then for every open $U \subset X$, the sequence $(x_{n_k})_k$ eventually stays in $U$. Arguing as above yields that $(x_n)_n$ converges in $X^*$.

Answer 2

I'll call $z = \infty_X$. Notice for instance that $\emptyset$ is open in $X$. This means that $\{z\}$ is open in $X^*$. What does this mean for sequences converging to $z$ in $X^*$?

On the other hand, if a convergent sequence $(x_n)$ in $X^*$ doesn't converge to $z$, then it is certianly not eventually the constant $z$. So, we can form the subsequence $(x'_n)$ described in (2). This new sequence also converges in $X^*$. Why does this sequence also converge in $X$? Hint: consider again the definition of open sets in $X^*$.