- #1
trap101
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Let A be a proper subset of X, and let B be a proper subset of Y .
If X and Y are connected, show that X × Y − A × B is connected.
Attempt: Proven before in my book, I know that since X and Y are each connected that X x Y is also connected. Keeping that fact in mind.
Pf: Assume (X x Y) - (A x B) has a separation.
--> (U u V) = (X x Y) - (A x B) where U and V are open disjoint nonempty subsets.
U and V can each be written in terms of their basis elements of the form (C x D) and (E x F) where C,E [itex]\subset[/itex] X and D,F [itex]\subset[/itex] V
therefore: (U u V) = (C x D) u (E x F)
= (C u E) x (D u F)
--> (C u E) [itex]\subset[/itex] X and (D u F)[itex]\subset[/itex]Y
Therefore I formed a separation between X and Y, but this cannot occur because we know
(X x Y) is connected --> Contradiction.
Now I feel something is a little messed up in the proof, but the idea is that I want to show that
(X x Y) has a separation which contradicts the fact I know about it.
Am I even close in my idea?
If X and Y are connected, show that X × Y − A × B is connected.
Attempt: Proven before in my book, I know that since X and Y are each connected that X x Y is also connected. Keeping that fact in mind.
Pf: Assume (X x Y) - (A x B) has a separation.
--> (U u V) = (X x Y) - (A x B) where U and V are open disjoint nonempty subsets.
U and V can each be written in terms of their basis elements of the form (C x D) and (E x F) where C,E [itex]\subset[/itex] X and D,F [itex]\subset[/itex] V
therefore: (U u V) = (C x D) u (E x F)
= (C u E) x (D u F)
--> (C u E) [itex]\subset[/itex] X and (D u F)[itex]\subset[/itex]Y
Therefore I formed a separation between X and Y, but this cannot occur because we know
(X x Y) is connected --> Contradiction.
Now I feel something is a little messed up in the proof, but the idea is that I want to show that
(X x Y) has a separation which contradicts the fact I know about it.
Am I even close in my idea?