Point set topology, hoemomorphism related questiosns

[elias001]

Hello

I am having difficulties in solving the following two questions.

1) For the first question, the author of the text states that if f:[a,b]-->R is a map, then I am f is a closed, bounded interval.

Question: Let X be subset of R, and X is the union of the open intervals (3n, 3n+1) and the points 3n+2, for n= 0,1,2,...
Let Y=(X-{2}) union {2}. Prove that there are continuous bijections f:X--->Y, g:Y--->X, but that X, Y are not homeomorphic.

I can create the bijection from X to Y, and Y to X.

From X to Y, I would map {2} to {1} and everything else gets map to itself, so I get both injective and surjective mapping.
From the Y to X direction, i would just map {1} to {2} and everything else gets map to itself. I get again a bijective mapping
But how do I show that the maps from X to Y and also from Y to X are both continuous.
X is composed of open intervals and singletons, likewise for the set Y.
Am I suppose to impose some sort of topology on X and Y and then described a basis elements. Also, why is the set I am f and I am g not counded or closed?

For the second problem:

Construct the homeomorphism f:[0,1]x[0,1]--->[0,1]x[0,1]
such that f maps [0,1]x{0,1} union {0}x[0,1] onto {0}x[0,1].

My difficulties with this question are several

first: Am i to interpret [0,1]x{0,1} union {0}x[0,1] to mean
([0,1]x{0,1}) union ({0}x[0,1]). If so then ([0,1]x{0,1}) union ({0}x[0,1])
is a subset of ({0,1] union (0})x({0,1} union [0,1]). By property of cartesian product. I am not sure how to proceed from here onwards.

Thank you in advance

 

[mighty2000]

for your help.

Hello,

Thank you for reaching out with your questions. I can offer some guidance on how to approach these problems.

For the first problem, you are correct in your approach of creating a bijection from X to Y and from Y to X. To show that these maps are continuous, you can define a topology on X and Y and then show that the inverse image of open sets is open. Since X and Y are composed of open intervals and singletons, you can define a topology on them by considering the open intervals and singletons as basis elements. Then, you can show that the inverse image of these basis elements under the bijections are open, thus proving the continuity of the maps.

As for why X and Y are not closed or bounded, it is because they are not closed intervals on the real line. X is a union of open intervals and singletons, and Y is the same as X but with the point 2 added. They are not closed intervals because they do not include their endpoints, and they are not bounded because they extend infinitely in both directions.

For the second problem, you are correct in interpreting [0,1]x{0,1} union {0}x[0,1] as ([0,1]x{0,1}) union ({0}x[0,1]), which can be written as ({0,1} union {0})x({0,1} union [0,1]). To construct the homeomorphism, you can consider the function f(x,y) = (x,y) for all (x,y) in [0,1]x[0,1]. This function maps [0,1]x{0,1} union {0}x[0,1] onto {0}x[0,1], and it is a continuous bijection. To show that it is a homeomorphism, you can use the same approach as in the first problem, by defining a topology on [0,1]x[0,1] and showing that the inverse image of open sets is open.

I hope this helps guide you in solving these problems. Let me know if you have any further questions.