Subspace topology

Give an example of a separable topological space possessing a non-separable subspace.

Prove the stated assertion. If $X$ is a topological space with topology $\Omega = \left\{ G\right\}$ and $E \subset X$, then the set $E$ is a topological space with the topology $\Omega_{E} = \left\{ G \cap E\right\} , G \in \Omega$. The topology $\Omega_{E}$ is called the topology induced by the topology $\Omega$ of the space $X$ on its subset $E$.

For $X \subset \mathbb {R}^{n}$, show that the usual topology on $X$ is the same as the subspace topology.

Proposition 1.2.2: If $A \subset X$ has the subspace topology, then $D \subset A$ is closed in $A$ iff $D = A \cap E$, where $E$ is closed in $X$.

One direction has already been shown: by the definition of the subspace topology, the open sets in $A$ are the intersections of $A$ with the open sets in $X$. Suppose $D$ is closed in $A$. Then $D = A \setminus F$, where $F$ is open in $A$. Then $F = A \cap G$, where $G$ is open in $X$. Hence, if $E = X \setminus G$, then $E$ is closed in $X$ and

Complete the proof by proving the converse: suppose $D = A \cap E$ where $E$ is closed in $X$, and show that $D$ is closed in $A$.

Suppose $A$ is a closed subset of $X$. Then $D \subset A$ is closed in $A$ (with the subspace topology) iff $D$ is closed in $X$.

Let $X$ be a metric space with metric $d$ and let $A$ be a subset of $X$. Show that the metric topology on $A$ given by $d$ is the same as the subspace topology.

Suppose $A \subset B \subset X$, and $B$ has the subspace topology. Show that, if $A$ is open in $B$ and $B$ is open in $X$, then $A$ is open in $X$.