Exercise 2.2.1 (Supersets generate the same relative topologies)

Question

Let $X$ be a topological space, let $S$ be a subspace of $X$, and let $E$ be a subset of $S$. Show that the relative topology that $E$ inherits from $S$ coincides with the relative topology that $E$ inherits from $X$.

Elu Thingol · Accepted Answer

Let $\mathcal{T}_X$ and $\mathcal{T}_S$ denote the topology on $X$ and the inherited topology on $S$ respectively. Similarly, let $\mathcal{T}_X\restriction_{E}$ and $\mathcal{T}_S\restriction_{E}$ denote the topologies on $E$ inherited from $X$ and $S$ respectively. We demonstrate that both topologies coincide.
\begin{proof}
\begin{description}
\item[$\implies$] Let $U \in \mathcal{T}_X\restriction_{E}$ be an open set from the topology of $E$ inherited from $X$. By definition, we can find an open set $V$ in the topology of $X$ such that $U = V \cap E$. Set $V^{\prime}:= V \cap S$. Then $V^{\prime}$ is an open set in $S$ by definition. But
$$U = V \cap E = V \cap (S \cap E) = (V \cap S) \cap E = V^{\prime} \cap E.$$
This, by definition, means that $U$ is an open set in a topology of $E$ inherited from $S$ as well, i.e., $U \in \mathcal{T}_S\restriction_{E}$.

\item[$\impliedby$] Let $U \in \mathcal{T}_S\restriction_{E}$ be an open set from the topology of $E$ inherited from $S$. By definition, we can find an open set $V^{\prime}$ in the topology of $S$ such that $U = V^{\prime} \cap E$. By the same definition, we can find an open set $V$ in the topology of $X$ such that $V^{\prime} = V \cap S$. But
$$U = V^{\prime} \cap E = (V \cap S) \cap E = V \cap (S \cap E) = V \cap E.$$
This, by definition, means that $U$ is an open set in a topology of $E$ inherited from $X$ as well, i.e., $U \in \mathcal{T}_X\restriction_{E}$.

\end{description}
\end{proof}

Exercise 2.2.1 (Supersets generate the same relative topologies)

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Exercise 2.2.1 (Supersets generate the same relative topologies)

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