# Neighborhood in Topological Subspace

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## 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 $\struct {S, \tau_S}$

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

## Proof

### 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:

- $D \cap S = \paren {V \cap S} \cup \paren {E \cap S} = U \cup E = E$

$\Box$

### 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 $\struct {S, \tau_S}$.

$\blacksquare$