It's the same question than in Prove that F $\subseteq$ R is closed if and only if every Cauchy sequences contained in F has a limit that is also an element of F.. Im following the same analysis book (Abott). I have a doubt with the left to right implication ($\implies$) proof provided by
the instructor manual. That's the following
Used definitions and theorems
Theorem 3.2.5 A point $x$ is a limit point of a set A if and only if $x = \lim a_n$ for some sequence $(a_n)$ contained in $A$ satisfying $a_n \neq x$ for all $n \in \mathbb{N}$
Definition 3.2.7 A set $F \subseteq \mathbb{R}$ is closed if it contains all its limit points.
Theorem to prove $F \subseteq {\rm I\!R}$ closed if and only if every Cauchy sequence contained in $F$ has a limit that is also an element of $F$
$(\implies)$
Instructor manual
Assume that the set $F \subseteq \mathbb{R}$ is closed. Then $F$ contains its limit points. We will show that that every Cauchy sequence $(a_n)$ contained in $F$ has its limit in $F$ by showing that the limit of $(a_n)$ is either a limit point or possibly an isolated point of $F$. Because $(a_n)$ is Cauchy, we know $\lim a_n$ exists. If $a_n \neq x$ for all $x$, then it follows from Theorem 3.2.5 that $x$ is a limit point of $F$. Now consider a Cauchy sequence $a_n$ where $a_n = x$ for some $n$. Because $(a_n) \subseteq F$ it follows that $x \in F$ as well. (Note that if an is eventually equal to $x$, then it may not be true that $x$ is a limit point of $F$.)
The problem I have with this proof is the line where it says
Now consider a Cauchy sequence $a_n$ where $a_n = x$ for some $n$.
Because $(a_n) \subseteq F$ it follows that $x \in F$ as well.
We can't apply theorem 3.2.5 ($\Longleftarrow$), because there are some values of the sequence where $a_n = x$
I would solve this issue the following way:
If $a_n = x$ happens a finite number of times I would create the subsequence $(a_{n_k})$ where the values $a_n = x$ are skipped. This subsequence converges to the same limit $x$, and it's possible to use theorem 3.2.5 to assert that it's a limit point thus $x \in F$.
Otherwise, if $a_n = x$ an infinite number of times we can take the subsequence $a_{n_k}$ where $a_{n_k} = x$ for all $k \in \mathbb{N}$, because $x \in (a_n) \subseteq F$ the limit $x \in F$.
Best Answer
You’re making it much harder than it really is: if $a_n=x$ for some $n$, then certainly $x\in F$, because by hypothesis $a_n\in F$. That’s it: you’re done. Remember, the goal is to prove that the limit $x$ is in $F$, and in this case there’s actually nothing to prove.
What you’ve done is not incorrect; it’s simply unnecessary.