Non empty perfect set and it's cardinality in different spaces

Question

Let $(X, \tau) $ be any topological space. $P\subset X$ is called perfect if $P'=P$ where $P'$ is the set of all limit ponits of $P$.

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1. If $(X, \tau) $ is a $T_1$ space, then any open set containing a limit point of $A$ contains an infinite subset of $A$. Hence existence of a limit point in a $T_1$ space makes the set infinite.

Corollary : A non empty perfect set in a $T_1$ space is infinite.

But the converse is not true. As $(X, \tau_{\text{indiscrete}}) $ with $|X|\ge \aleph_{0}$ is an example of a non $T_1$ space where every non empty perfect set (infact only non empty perfect set is $X$ itself) is infinite .

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2. Let $(X, d) $ be a complete metric space. Then any non empty perfect set in $(X, d) $ is uncountable.

As $\emptyset \neq P\subset X$ perfect set.

• $P$ closed and $X$ complete implies $(P, d_P) $ complete.

• $ P$ has no isolated points.

$(P, d_P) $ is complete metric space with no isolated point ( an application of Baire's theorem) implies $P$ is uncountable.

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3. Let $(X, \tau) $ be a locally compact Hausdorff space and $P$ be a non empty perfect set. Does this implies $P$ is uncountable?

Proof: $P\neq \emptyset$ implies $\exists x\in P$ .Since $(X, \tau)$ is Hausdorff space ( by $1$) , $P$ is an infinite set.

To show $P$ is not countably infinite . Suppose $P$ is countably infinite and $P=\{x_1,x_2,\ldots\}$.

Let $V_0$ is an open set containing $x_1$. Then by Local compactness of $(X,\tau)$ , $\exists V_1\subset V_0$ open set such that $x_1\in \overline{V_1}\subset V_0$ and since $x_1$ is a limit point of $P$ , $V_1\setminus \{x_1\} \cap P\neq \emptyset$ and $\overline{V_1}$ is compact.

Let $x_2\in V_1\setminus \{x_1\} \cap P$ Again by Local compactness, we can find an open set $V_2\subset V_1$ such that $x_2\in \overline{V_2}\subset V_1$ and $\overline{V_2}$ is compact. Again $V_2\setminus \{x_2\} \cap P\neq \emptyset$

Continuing in that way, we produce a sequence of open sets $(V_n) $ with the properties :

1. $\overline{V_{n+1}}\subset {V_n}$

2. $x_n\notin V_{n+1}$

3. $V_{n+1}\cap P\neq \emptyset$

Put $K_n=\overline{V_n}\cap P$

As $2$ implies $x_n \notin{V_{n+1}}$ implies $\bigcap_{n}K_n=\emptyset$.

But $(K_n) $ is a descending sequence of non empty compact sets , hence $\bigcap_{n}K_n\neq \emptyset$ which is a contradiction.

Consequence: A locally compact Hausdorff space with no isolated point in uncountable.

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Question: In a topological space $(X, \tau) $ every non empty compact sets are uncountable does this implies $(X, \tau) $ locally compact $T_2$ space?

An indiscrete topology on an uncountable set is locally compact but not Hausdorff where every non empty perfect set is uncountable.

Question 1 Does there exists any Hausdorff space which is not locally compact but every non empty perfect set is uncountable?

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Question2 : Let $(X, \tau) $ be a Baire space implies every $\emptyset \neq P$ perfect subset of $X$ is uncountable. Is the statement true? What about the converse?

Is all my above argument correct?

Edit: An infinite dimensional Banach space is not locally compact (as closed unit ball is not compact) . Hence $Q .1$ is now solved.