1) Fact: Let X be a metric space. Then the set X is(adsbygoogle = window.adsbygoogle || []).push({}); openin X.

Also, the empty set is open in X.

Why??

2) Let E={(x,y): x>0 and 0<y<1/x}.

By writing E as a intersection of sets, and using the following theorem, prove that E is open.

Theorem: Let X,Y be metric spaces. If f:X->Y is continuous on X, then f^{-1}(V) is open in X whenever V is open in Y.

Proof:

E = {(x,y): x>0} ∩ {(x,y): y>0} ∩ {(x,y): (1/x)-y>0}

A finite intersection of open sets is open, so it's enough to show each of the 3 sets are open.

To prove that {(x,y): (1/x)-y>0} is open, let f(x,y)=(1/x)-y.

Then {(x,y): (1/x)-y>0}=f^{-1}(0,∞) is open because the interval (0,∞) is open and f is continuous.

...

Now I don't understand why we can say that f is continuous and use the above theorem. When we break E into 3 sets, each set is to be treated separately and independently, so for {(x,y): (1/x)-y>0}, we don't assume x>0, but then f will not be continuous (f is not continuous at x=0), then how can we use the above theorem to prove that {(x,y): (1/x)-y>0} is open?? I don't understand...

Can someone please explain?

Thanks for any help!

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# Metric space & Open sets

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