Questions about open neighborhoods and topologies Statement:

For each $n \in \mathbb N$, let $C_n = \{(n, 1), (n, 2), (n, 3), (n, 4),\ldots\}$, and put
$X = \cup_{n\ge1}C_n, \ X^+ = X \cup \{(0, 0)\}$. Give $X^+$ a topology by declaring $G \subseteq X^+$ open if and only if: either ($0, 0) \not \in G$ or $\exists n_0 \in \mathbb N$ such that $\forall n \ge n_0, C_n \setminus G$ is finite. Prove that no sequence in $X$ can converge to $(0, 0)$.

Proof:

Suppose there is a $(z_i) \to (0, 0) \in X^+$ where $z_i = (x_i, y_i) \in X$. Then every subsequence of $(z_i)$ converges $(0, 0)$ also.


Case 1: One individual ‘column’ $C_n$ might include $z_i$ for infinitely many $i$. If
so, these $z_i$ would form a subsequence. And this subsequence fails to converge to $(0, 0)$ because $\{(0, 0)\} \cup C_{n+1} \cup C_{n+2} \cup C_{n+3} \cup \ldots$ is an open neighbourhood of $(0, 0)$ which the subsequence fails to enter. Contradiction!


Case 2: Otherwise, put $N = X^+ \setminus \text{(the range of the sequence $(z_i)$)}$, and this set misses only a finite number of terms in each column, so it is an open
neighbourhood of $(0, 0)$ which the sequence fails to enter. Contradiction.

My questions about the proof above:

*

*How is this proof broken into cases explicitly or how do they construct the sequences explicitly? In the second case we take finite number of elements from each column to construct a subsequence and in the first one we take infinite number of elements from each column to construct a subsequence, but WLOG consider only one. Is that correct?


*How do we know $\{(0, 0)\} \cup C_{n+1} \cup C_{n+2} \cup C_{n+3} \cup \ldots$ is an open neighbourhood of $(0, 0)$? We know $\{(0, 0)\} \not \in C_{n+i}$ for $i \ge 1$ and so $\cup_{i \ge 1} C_{n+i}$ is open. But is $(\cup_{i \ge 1} C_{n+i}) \cup \{(0, 0)\}$ open, too? How?


*Where in the proof do we use the definition of topology given in the statement?
 A: Regarding your first question: It is almost correct what you are saying: For case 1, we are only considering one column $C_n$ with infinitely many $z_i$ contained in it. It's important to note that we are not explicitly constructing any sequences. It is sometimes simply more useful to us that these (sub-)sequences exist. Then it becomes irrelevant to know what exactly they look like. I'll give an overview of the proof:
For case 1, let's assume (for now) that $n=1$, i.e. the column $C_1$ includes infinitely many terms $z_i$. Let $I := \{ i \in \mathbb{N} \mid z_i \in C_1 \}$. Then consider $(z_i)_{i \in I}$. We indeed have
$$
\{ z_i \mid i \in I \} \cap \big(\{(0,0)\} \cup C_2 \cup C_3 \cup C_4 \cup \ldots\big) = \emptyset.
$$
The set $\{(0,0)\} \cup C_2 \cup C_3 \cup C_4 \cup \ldots$ is an open neighborhood of $(0,0)$, since $C_n \setminus G = \emptyset$ for $n\geq n_0 := 2$, so in particular $C_n \setminus G$ is finite. Here we directly use the definition of the topology given in the statement. Then it is not hard to see that in the general case
$$
\{(0,0)\} \cup \bigcup_{i\geq1}C_{n+i}
$$
is an open neighborhood of $(0,0)$.
For case 2, we consequently assume that all columns $C_n$ contain only finitely many terms $z_i$. Let $Y$ denote the range of $(z_i)_{i \in \mathbb{N}}$. By your definition of $N$, we see that $$C_n \setminus N = C_n \cap Y,$$ so by our assumption $C_n \setminus N$ is finite for all $n \geq n_0 := 1$, and hence $N$ is an open neighborhood of $(0,0)$, again by the definition of the topology.
I hope this was helpful!
