About proof of theorem on topology

Question

Theorem. If a convergent sequence in a metric space has infinitely many distinct points, then its limit is a limit point of the set of points of the sequence.

In the previous image is the proof of a theorem, may question is about the information used to reach absurd, is told that since x is in the sequence, the sequence should have only finitely many distinct a points, but this necessarily true? If we have like all {$x_n$} being distinct a points, for example n being negative integers, and all {$x_n$} being x, for all n being natural numbers, this is a convergent serie with limit x, and with infinitely many diferentes points, even do the aforementioned proof keep being true?

How may I proof it?

Answer 1

We have a sequence $(x_n)_n$ with limit $x$. This is a function, that assigns to each $n$ some point $x_n \in X$. The set of points is $A = \{x_n: n \in \mathbb{N}\}$, which is a different object, which can be finite. $x$ is not in the sequence per se, nor need it be in $A$.

But suppose $x$ is not a limit point of $A$ (the term limit point suggests $x$ should be a limit point of $A$, and this is very often the case, except in a trivial case we will see now). Then there is some neighbourhood $S_\epsilon(x)$ of $x$ such that $S_\epsilon(x) \cap A = \{x\}$ or $\emptyset$. We can apply the definition of a convergent sequence for this $\epsilon$, and we get some $N$ such that for all $n > N$, we have that $x_n \in S_\epsilon(x)$. All such points are in $A$, so for all those $n > N$: $x_n = x$. So the sequence is a trivial convergent sequence: all its tail consists of values $x$. And so $A$ can be at most $\{x_!,x_2,\ldots,x_N, x\}$ and $A$ is finite. A finite set actually never has a limit point in a metric space. So $x$ is not a limit point of $A$ iff $A$ is finite.

Answer 2

A limit point (or accumulation point) for a subset $U$ of the metric space $X$ is $x$ such that, for every $\varepsilon>0$, there exists $u\in U$ such that $u\ne x$ and $u\in S_\varepsilon(x)$.

This is the same as saying that, for every $\varepsilon>0$, $S_\varepsilon(x)$ contains infinitely many points of $U$ (prove it).

So, if $x$ is not a limit point of the support $U$ of the sequence $\{x_n\}$ (that is, the set of its values), there exists $\varepsilon>0$ such that $S_\varepsilon(x)$ contains at most finitely many points of $U$ different from $x$. Taking the minimum distance of these points from $x$, we see that we can assume $S_\varepsilon(x)$ contains no point of $U$ different from $x$.

However, the sequence converges to $x$, so there exists $k$ such that, for $n>k$, $x_n\in S_\varepsilon(x)$. Therefore, for $n>k$, we have $x_n=x$ and so $$ U=\{x_0,x_1,\dots,x_k,x\} $$ is finite.

Answer 3

(I). An equivalent, and often useful, way of saying that $(x_n)_n$ converges to $x$ in a metric space $(X,d)$ is that for every nbhd $U$ of $x,$ the set $\{n:x_n\not \in U\}$ is finite:

(i). If $(x_n)_n$ converges to $x,$ and $U$ is a nbhd of $x,$ then $B_d(x,r)\subset U$ for some $r>0.$ And there exists $n_0$ such that $\forall n>n_0\; ( d(x,x_n)<r) .$ So $$\{n:x_n\not \in U\}\subset \{n:x_n \not \in B_d(x,r)\subset \{n:n\leq n_0\}.$$

(ii). If $\{n: x_n\not \in U\}$ is finite for every nbhd $U$ of $x$ then for every $r>0$ the finite set $\{n:x_n\not \in B_d(x,r)\}$ is a subset of $\{n:n\leq n_0\}$ for some $n_0\in \mathbb N.$ So there exists $n_0$ such that $\forall n>n_0 \;(x_n\in B_d(x,r)).$ Equivalently there exists $n_0\in \mathbb N $ such that $\forall n>n_0\;(d(x,x_n)<r).$

(II). Let $(x_n)_n$ converge to $x,$ such that $A=\{x_n: n\in \mathbb N\}$ is infinite.

If $U$ is any nbhd of $x$ then $\{n:x_n\not \in U\}$ is finite, so take $n_0$ such that $\forall n>n_0\;(x_n\in U).$

Since $A$ is infinite and $B=\{x_n:n\leq n_0\}$ is finite, the set $A$ \ $B=\{x_n:n>n_0\}$ is infinite so there exists some $n_1>n_0$ with $x_{n_1}\ne x.$ But $x_{n_1}\in \{x_n:n>n_0\}\subset U.$ Therefore every nbhd $U$ of $x$ contains a member of $A$ that is not equal to $x. $ Equivalently, $x$ belongs to the closure of $\{x_n:x_n\ne x\}.$