Undergrad Problem about a connected subspace

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The discussion centers on a potential error in a topology exercise from Munkres, which claims that the space (X×Y) - (A×B) is connected when X and Y are connected, and A and B are proper subsets. A counterexample involving the removal of the product (0,1) x (0,1) from R² is presented to challenge this assertion. The argument highlights that subsets of the form {a} x Y and X x {b} remain connected, supporting the claim that the union of connected sets that intersect is also connected. Ultimately, the initial concern about the counterexample is resolved, affirming that the original statement about connected products holds true. The discussion concludes with an acknowledgment of the misunderstanding regarding the proof of connectedness in product spaces.
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Helo, I believe that the folowing exercise from Topology by Munkres is incorrect:
"Let A be a proper subset of X, and let B be a proper subsert of Y. If X and Y are conected, show that
##(X\times Y)-(A\times B)## is connected"
I think I can prove it wrong however I'm not sure and would like to discuss it.
 
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facenian said:
Helo, I believe that the folowing exercise from Topology by Munkres is incorrect:
"Let A be a proper subset of X, and let B be a proper subsert of Y. If X and Y are conected, show that
##(X\times Y)-(A\times B)## is connected"
I think I can prove it wrong however I'm not sure and would like to discuss it.

Then give a counterexample.
 
Math_QED said:
Then give a counterexample.
I think I have it, however I don't want to mislead someone else to a probably mistaken answer that's why I would like to know if anybody already now the solution or has an opinion.
I think I must be wrong otherwise there is a mistake in Munkres' book
 
Let's just look in the (x,y) plane R^2, and remove the product (0,1) x (0,1). Do you see how to connect up all points of the complement by drawing vertical and horizontal lines? The same thing works in general. try it yourself after removing from the plane say all points having both coordinates rational. I.e. remove QxQ from RxR. The point is in general that in the product space XxY, all subsets of the form {a} x Y and Xx{b} are connected, (and meet at (a,b)), and the union of two connected sets that meet is also connected. Does that do it? I have not written anything down.
 
mathwonk said:
Let's just look in the (x,y) plane R^2, and remove the product (0,1) x (0,1). Do you see how to connect up all points of the complement by drawing vertical and horizontal lines? The same thing works in general. try it yourself after removing from the plane say all points having both coordinates rational. I.e. remove QxQ from RxR. The point is in general that in the product space XxY, all subsets of the form {a} x Y and Xx{b} are connected, (and meet at (a,b)), and the union of two connected sets that meet is also connected. Does that do it? I have not written anything down.

Thank you, now a see the mistake in the counterexample I thought I had.
 
in fact this is almost the same as the proof that a product of connected spaces is connected.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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