Intersection of Neighborhoods in Topological Space is Neighborhood

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $x \in S$.

Let $M, N$ be a neighborhoods of $x$ in $T$.


Then $M \cap N$ is a neighborhood of $x$ in $T$.


That is:

$\forall x \in S: \forall M, N \in \mathcal N_x: M \cap N \in N_x$

where $\mathcal N_x$ is the neighborhood filter of $x$.


Proof

By definition of neighborhood:

$\exists U_1 \in \tau: x \in U_1 \subseteq M$

where $U_1$ is an open set of $T$.

$\exists U_2 \in \tau: x \in U_2 \subseteq N$

where $U_2$ is an open set of $T$.

Thus by Set Intersection Preserves Subsets:

$U \subseteq M \cap N$

where $U = U_1 \cap U_2$

The result follows by definition of neighborhood of $a$.

$\blacksquare$


Also see


Sources