here is the one of the answer my question in the cite but I would like to ask something,
'If $(X,d)$ is a complete metric space, and $A$ is perfect and $x \in A$ it follows that for any $r>0$, $D(x,r) \cap A$ is non-empty (where $D(x,r) = \{y \in X: d(x,y) \le r\}$ is the closed ball of radius $r$ around $x$).
$D(x,r)$ is closed and hence so is $D(x,r) \cap A$ and thus also complete.
So if $D(x,r) \cap A$ were countable for some $r>0$, we must have an isolated point $a_0 \in D(x,r) \cap A$ (or we'd contradict the Baire property) and this can only happen if $A$ already had an isolated point, which is not the case. So all points of $A$ are condensation points.'
so in this solution every closed ball of radius $r$ is taken and intersection of that closed balls with A is uncountable, but can we conclude it $x$ is at the same time condensation point of the perfect set $A$? Because of the definition of condensation point is the one whose all OPEN neighbourhoods are uncountable , can we conclude from this solutıon every open ones uncountable? if so, can we say that $a_{0}$ is also isolated point of $A$ itself?
any hint would be great, thanks in advance.
Best Answer
Let $B(x,r)$ be the open ball of radius $r$ centred at $x$. Clearly
$$D\left(x,\frac{r}2\right)\subseteq B(x,r)$$
for every $r>0$, so if some $B(x,r)$ were countable, $D\left(x,\frac{r}2\right)$ would also be countable, and $x$ would be an isolated point by the Baire category argument mentioned in the question.