- #1
kingwinner
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1) Fact: Let X be a metric space. Then the set X is open in 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!
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!