Proving [a,b) is first countable but not second countable

Question

Since this half-open interval is neither open nor closed, I have a problem with proving the theorem. Can I have an answer ?

Answer 1

Let $(X,\tau)$ be a first countable space, we prove any subspace $Y$ is also first countable.

Take $y\in Y$, we must find a countable local basis for $y$. Let $\mathcal B$ be a local basis for $y$ in $X$. Consider the set $\mathcal B'=\{B\cap Y | B\in \mathcal B\}$

We must prove this is a local basis for $y$ in $Y$. So take an open set $U'$ in $Y$ containing $y$. We must have that $U'=U\cap Y$ such that $U$ is an open set in $X$ containing $y$. Therefore there exists $B\in \mathcal B$ such that $y\in B$ and $B\subseteq U$. We conclude that $y\in B\cap Y\subseteq U'$, and clearly $B\cap U\in \mathcal B'$. Therefore $\mathcal B'$ is a local basis for $y$ in $Y$. Since we can find a countable local basis for each point in $y$ we conclude $Y$ is first countable.

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Let $(X,\tau)$ be a second countable space, we prove any subspace $Y$ is also second countable.

Let $\mathcal B$ be a basis for $X$. Consider the set $\mathcal B'=\{B\cap Y | B\in \mathcal B\}$

We must prove this is a basis for $y$ in $Y$. So take an open set $U'$ in $Y$ and a point $u\in U$. We must have that $U'=U\cap Y$ such that $U$ is an open set in $X$ containing $u$. Therefore there exists $B\in \mathcal B$ such that $u\in B$ and $B\subseteq U$. We conclude that $u\in B\cap Y\subseteq U'$, and clearly $B\cap U\in \mathcal B'$. Therefore $\mathcal B'$ is a basis for $Y$. Since we can find a countable basis $Y$ we conclude $Y$ is second countable.