compactnessgeneral-topology
I would like to see an example of a Hausdorff compact space $X$ for which there exists an uncountable set $A \subseteq X$ such that:
This question came to my mind when I was thinking about the compact interval $[0,1]$ and its subset $\big\lbrace \frac1n ; \, n \in \mathbb{N} \big\rbrace$. This set satisfies the two conditions above, but it is countable.
Suppose $X$ is a locally compact Hausdorff space and $Y \in \Gamma(X)$. I'll prove that $Y$ is a locally compact Hausdorff space with the subspace topology.
I
Suppose $x,y \in Y$ and $x \not= y$.
Hence $Y$ is Hausdorff.
II
Let $x \in Y$. We have:
$\;\;\;\;\exists V \in \Gamma(X), \{x\} \subset V \ \subset cl_X(V) \subset Y \ \text{ such that } cl_X(V)\text{ is $X-$compact.}$
$cl_X(V)$ is $Y-$compact as well. Also $V \in \Gamma(Y)$
As $cl_X(V) = cl_X(V) \cap Y$, $cl_X(V)$ is $Y-$closed.
By 2 and 4, $cl_Y(V) \subset cl_X(V)$.
As closed subsets of compact sets are compact,$cl_Y(V)$ is $Y-$compact.
Now we have all the ingredients ready. We have:
Best Answer
Let $X=[0,1]^\Bbb R$ (with the product topology, of course). Let $A$ be the set of all elements $x\in X$ such that $x$ takes the value 1 in exactly one coordinate and the value 0 in all other coordinates. It is easy to verify (if you know Tychonoff's Theorem) that this does what you ask.