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Proof it is a base in a metric space...

Lolyta

New member
Feb 21, 2012
4
Hello. This is my problem:
Let (X,T) a metric space, if $A\subset X$ is dense, proof that $B=$ { $B_d$(a,$\frac{1}{n}$) :$a\in A$ , $n\in N$ } is base for T. ($B_d$ is a ball).
I don't know how to do it: but here i let you two propositions I think they could be useful (but I don't know how to use them well).

a) B is base if and only if $\forall U \in T$ with U open and $\forall x \in U$ exists $b\in B$: $x\in B \subset U$
b) B is base if we can put every open as an union of elements of the base.

Thank you so much for any help!
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Hello. This is my problem:
Let (X,T) a metric space, if $A\subset X$ is dense, proof that $B=$ { $B_d$(a,$\frac{1}{n}$) :$a\in A$ , $n\in N$ } is base for T. ($B_d$ is a ball).
If $\alpha>0$ and $x\in B(a,\alpha)$ then there is a $\beta>0$ such that $B(x,\beta)\subset B(a;\alpha)$
You should have proved that basic theorem before.

Now just note if $t\in U$ the there is a ball $B(t;\alpha)\subseteq U$ which must contain a point $a\in A$.

Now you are just about done.
 

Lolyta

New member
Feb 21, 2012
4
If $\alpha>0$ and $x\in B(a,\alpha)$ then there is a $\beta>0$ such that $B(x,\beta)\subset B(a;\alpha)$
You should have proved that basic theorem before.

Now just note if $t\in U$ the there is a ball $B(t;\alpha)\subseteq U$ which must contain a point $a\in A$.

Now you are just about done.
do you use A is dense? where?
Thank you so much!
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
do you use A is dense? where?
Thank you so much!
The fact that $A$ is dense in $X$ means that any open set in $X$ contains a point of $A$.