Analysis
$f:A\rightarrow B$
$f$ is continuous iff when ever F is closed in B we have that the preimage of F is closed in A

Question

Analysis
$f:A\rightarrow B$ 
$f$ is continuous iff when ever F is closed in B we have that the preimage of F is closed in A

zzr0ck3r · Answer

A and B are metric spaces, and I have the fact that this is true for the exact same thing except with open sets. I also only need converse way. So assume when F is closed in B we have its preimage is closed in X, then show f is continuous

hartnn · Answer

lol, sorry, this is above me :P

zzr0ck3r · Answer

word

nincompoop · Answer

the language is a bit discombobulating laughing out loud

zzr0ck3r · Answer

?

nincompoop · Answer

a fancy way of reiterating what hartnn said

zzr0ck3r · Answer

ahh word, well tnx for reading:)

zzr0ck3r · Answer

ok figured it out, just in case you want to know...

assume Y is a open set in B, then B\Y is closed in B
thus the primage is closed by assumption
so
x\f^-1(B) is closed, so f^-1(B) is open, and thus f is continuous