second countable topological space and relative topology

Question

Excuse me can you see the following question .. Let X be a seconed countable topological space and Let A be an uncountable subset of X . Prove that the topology induced on A is not the discrete topology >>>

i tried on it using the contrary by assuming that A is uncountable discrete topological space and get that it is not seconed countable , in fact it is not separable >> but i deal with it as if X is not separable then A induced topological space is not separable is it true ??

Answer 1

If X were discrete, then X would be second countable because every base includes all the singletons.
As an uncountable set includes an uncountable number of singletons, it cannot be discrete.

Answer 2

By definition, a second-countable space has a countable basis. If there were an uncountable subspace with its induced topology the discrete topology, then there would exist an uncountable collection of singleton open sets in the ambient space, and hence uncountably pairs of points. The intersection of the sets in any of these pairs is empty. I'd recommend seeing if you can finish the proof from there. If not, I can give more hints. It's actually an interesting result, in my opinion at least, since it is rather unintuitive that such there is such a weak condition, that, were it to be false, imply that any uncountable space has the discrete topology - which would so greatly simplify the study of topology.

Answer 3

If $X$ is second countable then any subspace $A$ of it is also second countable (take the intersections of the members of the countable base for $X$ with $A$, these are open in $A$ and form a countable base for $A$).

If $A$ has the discrete topology, then any base for it must contain all sets $\{a\}$ (it's the only way we can write these open sets in $A$ as a union), and so has size at least $|A|$.

These statements combined give the result.