Question: Suppose $(X,d)$ is a metric space and $\{x_n\},\{y_n\}$ are sequences in $X$. $\{x_n\}$ has no convergent sub-sequence in $X$ and $\{y_n\}$ has only one convergent subsequence in $X$:- $\{y_{m_n}\}$ . Let $S$ be a subset of $X$.
Then,which of the following is true:
$1$. If $z \in S$ is an isolated point of $S$, then it is possible that $z$ may be a boundary point of $S$.
$2.$ Every subset $H$ of $\{x_n\}$ consists entirely of isolated points of $H$ and is hence open in $X$.
$3.$ Every subset $H$ of $\{x_n\}$ consists entirely of isolated points of $H$ and is hence closed in $X$.
$4.$ The set $\{y_{m_n}\}$ is closed in $\{y_n\}$.
$5.$ The set $\{y_n\} - \{y_{m_n}\}$ is open in $X$.
Attempt:
$1.$ True. An isolated point in $S$ may be at $0$ distance from $S^c$. Example: the set of Natural numbers in Real Numbers.
$2.$ False. Every subset $H$ of $\{x_n\}$ consists entirely of isolated points of $H$ and each isolated point $i$ is open in $H$ and so is their union. . But, $\bigcup i$ may not be isolated in $X$ and hence may not be open. ( As then, the set itself becomes a part of the boundary of the set).
$3.$ True. Since, Every subset $H$ of $\{x_n\}$ consists entirely of isolated points in $X$, they possess no accumulation point in $X$. Therefore, no element from $H^c$ can be at $0$ distance from $H$ and hence, cannot be a part of the boundary. Thus, the boundary is included in $H$ itself and is hence closed in $X$.
$4.$ False. The limit point of $\{y_{m_n}\}$ may not belong to $\{y_{m_n}\}$. Therefore, it may not be closed.
$5.$ False. The set $\{y_n\} - \{y_{m_n}\}$ consists entirely of isolated points of $y_n$ since $y_n$ has only one convergent sub-sequence. From point $3$, these set of isolated points are closed in $X$.
Am I correct in my attempt? Thanks a lot for your help.
I'm not convinced by your answer to 4 or 5. Here's why:
Claim: $(y_n)_n$ is constant.
Proof: If $\# \{y_{m_n}\}_n \geq 2$, then we can create a different convergent subsequence of $(y_n)$ by simply taking out all the terms $(y_{m_n})_n$ which have a particular value, provided that there are infinitely many terms left over. (Such a value is guaranteed by the infinite pigeonhole principle).
Likewise, if there is a value $a = y_N \in \{y_{n}\}\backslash \{y_{m_n}\}$, then adding on this $N$th term to $(y_{m_n})$ yields another distinct convergent sequence! $\qquad\square$
From this, it's clear that
4) $\{y_{m_n}\}=\{\ast\}=\{y_{n}\}$ is open in $\{y_{n}\}$, and
5) $\{y_{n}\}\backslash \{y_{m_n}\}=\emptyset$ is open in $X$.
$ $
Otherwise, I have no problems!