How $A=\{x_n:n\in \mathbb N\}$ is an infinite set?Can you explain the proof?

Question

Can you please explain the underlined arguments? How does it deduce that $A$ is infinite? I understood that when $y\in X$. There is an $U\in \mathscr O$ such that $x\in U$(since $\mathscr O$ is an open cover for $X$). Since $U$ is open there is an open ball $B_d(y,\epsilon)\subseteq U$. By Archimedean property I can deduce $B_d(x,\frac{1}{n})\subseteq U, \forall n>N$ How $x_n\neq y$ and $A$ is infinite?

Answer 1

We have

• $B_d(x_n, \frac{1}{n}) ⊈ U$ for every $n$,
• $B_d(y, \frac{1}{n}) ⊆ U$ for every $n > N$.

Therefore, so $x_n ≠ y$ unless $n ≤ N$.

Answer 2

Here, you prove that for each $y \in X$, there exists $N$ such that for $n > N$, one has $x_n \neq y$.

This is sufficient to ensure that $A$ is infinite.

Answer 3

Here is an alternative argument. Assume that $A$ is finite. Then there must exist some $a \in A$ such that $x_n = a$ for infinitely many $n$. There exist $U \in \mathcal O$ such that $a \in U$ and $\varepsilon > 0$ such that $B_d(a,\varepsilon) \subset U$. Take any $n$ such that $x_n = a$ and $\frac{1}{n} < \varepsilon$. Then $B_d(x_n,\frac{1}{n}) \subset U$ which is impossible.