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#### bw0young0math

##### New member

- Jun 14, 2013

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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?

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?

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