# metric spaces

• Oct 28th 2010, 03:06 PM
Godisgood
metric spaces
Let (M,d) be a metric space.
1. Choose x element of M. Show that a set U subset of M is a neighbourhood of x if and only if x is an element of the Interior of U
2. Prove that the intersection U intersection V of any two neighbourhoods U and V of a point x element of M is also a neighbourhood of x.
• Oct 28th 2010, 10:47 PM
Drexel28
Quote:

Originally Posted by Godisgood
Let (M,d) be a metric space.
1. Choose x element of M. Show that a set U subset of M is a neighbourhood of x if and only if x is an element of the Interior of U
2. Prove that the intersection U intersection V of any two neighbourhoods U and V of a point x element of M is also a neighbourhood of x.

Thanks soo much. I am really struggling with this since it is the first time I am taking a course of this nature, so please go in steps
Thanks for the help

What is a neighborhood? Do you take the Bourbaki "set containing an open set containing the point" or just "open set containing point"
• Oct 30th 2010, 11:08 AM
Plato
Quote:

Originally Posted by Drexel28
What is a neighborhood? Do you take the Bourbaki "set containing an open set containing the point" or just "open set containing point"

@Godisgood,
O.K. then if $O$ is an open set and $a\in\O\subseteq V$ the by definition $a\in\mathscr{I}(V)$, interior of $V$.
On the other,hand, if $b\in\mathscr{I}(V)$ the by definition of interior, an open set $\left( {\exists Q} \right)$ such that $b\in Q\subseteq V$ so $V$ is a neighborhood of $b$.
If each of $V~\&~U$ is a neighborhood of $c$ there are open sets $c \in O \subseteq V\;\& \;c \in Q \subseteq U$.
But $O\cap Q$ is open and $c\in O\cap Q \subseteq V\cap U$.