- #1
seyma
- 8
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Let (X,d) be a metric space A and B nonempty subsets of X and A is open. Show:
A[tex]\cap[/tex]B = [tex]\oslash[/tex] Iff A[tex]\cap[/tex]B(closure)= empty
Only B closure
it is easy to show rigth to left but how can i use A's open property I try to solve with contradiction s.t. there exist r>0 Br(p)[tex]\subseteq[/tex]A[tex]\cap[/tex]B(closure) but i cannot come conlusion. Can you help me please ? :(
A[tex]\cap[/tex]B = [tex]\oslash[/tex] Iff A[tex]\cap[/tex]B(closure)= empty
Only B closure
it is easy to show rigth to left but how can i use A's open property I try to solve with contradiction s.t. there exist r>0 Br(p)[tex]\subseteq[/tex]A[tex]\cap[/tex]B(closure) but i cannot come conlusion. Can you help me please ? :(
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