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[SOLVED] Graph is closed iff f cont. (using compact)

bw0young0math

New member
Jun 14, 2013
27
X:T2,Y:T2&compact.
f:X->Y : continous iff G(f)={(x,f(x))/ x∈X}:closed in XxY.

the proof of it is written in my book like below.

←Claim: f: cont. at arbitrary x∈X.
Let H: arbitrarary open neiborhood of f(x).
Arbitrary y(≠f(x))∈Y, (x,y)∈G(f)^c. Thus there exist U_y: open nbd(=neiborhood) of x, V_y: open nbd of y s.t. (x,y) ∈U_y x V_y ⊂G(f)^c.
Let A={H}∪{V_y/y∈Y-{f(x)}}: open cover of Y. Since Y: compact, {H,V_y1,V_y2,...,V_y} subcover of A.
Let U=U_y1∩U_y2∩...∩U_yn, V=V_y1∪V_y2∪...∪V_yn.
Then UxV ∩ G(f)=0(empty set.). Thus x∈U⊂f^-1(H). Therefore f: cont. at x.


However, I don't know Thus x∈U⊂f^-1(H). in last sentence.

Many books expain the property using projection map&close map. However, my book wrote the proof above. Is the proof in my book right?
 
Last edited:

hmmm16

Member
Feb 25, 2012
31
Re: graph is closed iff f cont. (using compact)

I think something has gone wrong with your posting of this question (unless something is wrong with my computer)
 

bw0young0math

New member
Jun 14, 2013
27
Re: graph is closed iff f cont. (using compact)

I think something has gone wrong with your posting of this question (unless something is wrong with my computer)
Thanks! I edited my writing.(heart):p
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Re: graph is closed iff f cont. (using compact)

Then UxV ∩ G(f)=0(empty set.). Thus x∈U⊂f^-1(H).
Suppose $x'\in U$. We need to show that $f(x')\in H$. Since $f(x')\in Y$ and $H,V_{y_1},\dots,V_{y_n}$ form a covering of $Y$, we have $f(x')\in H$ or $f(x')\in V_{y_i}$ for some $i$. However, in the second case $G(f)\ni(x',f(x'))\in U\times V_{y_i}\subseteq U\times V$, and $(U\times V)\cap G(f)=\emptyset$, a contradiction.