Neighborhood in Topological Subspace

Theorem
Let $\struct{X, \tau}$ be a topological space.

Let $S \subseteq X$ be a subset of $X$.

Let $\tau_S$ denote the subspace topology on $S$.

Let $x \in S$ be an arbitrary point of $S$.

Let $E \subseteq S$.

Then:
 * $E$ is a neighborhood of $x$ in $\left({S, \tau_S}\right)$


 * $\exists D \subseteq X$ such that:
 * $D$ is a neighborhood of $x$ in $X$
 * $E = D \cap S$.
 * $E = D \cap S$.

Necessary Condition
Let $E$ be a neighborhood of $x$ in $\struct{S, \tau_S}$.

By the definition of neighborhood:
 * $\exists U \in \tau_S : x \in U \subseteq E$

Now, by the definition of the subspace topology:
 * $\exists V \in \tau: U = V \cap S$

Take $D := V \cup E$.

We have that:
 * $V \subseteq D$

and:
 * $V \in \tau$

Thus $D$ is a neighborhood of $x$ in $X$.

Therefore it holds true that:
 * $D \cap S = \paren{V \cap S} \cup \paren{E \cap S} = U \cup E = E$

Sufficient Condition
Let $\exists D \subseteq X$ such that $D$ is a neighborhood of $x$ in $\struct{X, \tau}$.

Let $E = D \cap S$

Then, by the definition of neighborhood:
 * $\exists V \in \tau: x \in V \subseteq D$

By the definition of the subspace topology:
 * $V \cap S \in \tau_S$

As a consequence:
 * $E = D \cap S \supseteq V \cap S \in \tau_S$

That is:
 * $E$ is a neighborhood of $x$ in $\left({S, \tau_S}\right)$.

Also see

 * Closed Set in Topological Subspace