Let $\tau$ and $\sigma$ be topologies over a set $X$ such that $\tau \subseteq \sigma$.
a) If $(X, \tau )$ satisfies a certain countability axiom then $(X, \sigma)$ also satisfy?
b) If $(X, \sigma)$ satisfies a certain countability axiom then $(X, \tau )$ also satisfy?
1.First-Countable
A topological space $T=(S,\tau)$ satisfies the First Axiom of Countability if and only if every point in $S$ has a countable local basis.
2.Second-Countable
A topological space $T=(S,\tau)$ satisfies the Second Axiom of Countability if and only if its topology has a countable basis.
3.Separable
A topological space $T=(S,\tau)$ is separable if and only if there exists a countable subset of S which is everywhere dense in $T$.
4.Lindelöf
A topological space $T=(S,\tau)$ is a Lindelöf if and only if every open cover of S has a countable subcover.
As 1,2,3 and 4 depend on countable sets, we can guarantee that if $\tau \subseteq \sigma$ and $\sigma$ is countable then $\tau$ is countable. But if $\tau $ is countable, nothing can be said of $\sigma$ so a) it can't be true, right?
b) 1. By the definition of first-countable, every point in $\sigma$ has a countable local basis in $T$.
So the intersection of $\tau$ with the countable local basis of $\sigma$ give us a countable local basis for $\tau$.But every point in $\tau$ is also a point in $\sigma$, so $\tau$ is also first countable.
2. By the definition of second-countable $\sigma$ has a countable basis.So, $\exists \cal{B} \subseteq \sigma$ $ then for all $U \in \sigma$, $U$ is a union of sets from $\cal{B}$.But \cal{B} is countable. As $\tau \subseteq \sigma$ it follows that a $\sigma$ itself is a union of sets from $\cal{B}$.So $\tau$ is also second countable.
3. Let $\mathbb R_7$, which is both separable ({7} is dense) Let $Y = \mathbb R$ \ {$7$} $\subseteq \mathbb R_7$. This space is not separable.
4. Let X be the space of all uncountable ordinals, together with the first uncountable ordinal itself,so the space $\sigma + 1$ is Lindelöf, but the subspace $\tau = \sigma + 1 - 1$ is not.
How is my answer?
Thanks.