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I am having difficulties in solving the following two questions.

1) For the first question, the author of the text states that if $f:[a,b]\rightarrow R$ is a map, then $\text{Im} f$ is a closed, bounded interval.

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

I can create the bijection from $X\text{ to }Y$, and $Y\text{ to }X$.

From $X\text{ to }Y$, I would map $\{2\}\text{ to }\{1\}$ and everything else would get mapped to itself, so I get both an injective and surjective mapping. From the $Y\text{ to }X$ direction, I would just map $\{1\}\text{ to }\{2\}$ and everything else would get mapped to itself. I get again a bijective mapping But how do I show that map from $X\text{ to }$Y and also $Y\text{ to }X$ 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 $X\text{ and }Y$ and then describe the basis elements? Also, why are the sets $\text{Im }F\text{ and Im }g$ not bounded or closed?

For the second problem:

Construct the homeomorphism $f:[0,1]\times[0,1]\rightarrow[0,1]\times[0,1]$ such that $f$ maps $[0,1]\times\{0,1\} \cup \{0\}\times[0,1]$ onto $\{0\}\times[0,1]$.

My difficulties with this question are:

Am I to interpret $[0,1]\times\{0,1\} \cup \{0\}\times[0,1]$ to mean $([0,1]\times\{0,1\}) \cup (\{0\}\times[0,1])$? If so, then $([0,1]\times\{0,1\}) \cup (\{0\}\times[0,1])$ is a subset of $([0,1] \cup \{0\})\times(\{0,1\} \cup [0,1])$, by a property of of the cartesian product. I am not sure how to proceed from here onwards.

Thank you in advance

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    I may be misunderstanding something, but how is $Y=(X-\{2\})\cup \{2\}$ different from $X$? Or do you mean $Y$ is $X$ shifted to the left by $2$ united with $\{2\}$?2012-09-14
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    I did the best I could to format your question into the expected format for this group. If you'd like an introduction to the markup language we use here, see [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). @tomasz just asked a question I had, so I can skip it. In the penultimate paragraph you asked "Am I to interpret..." My answer is yes: the usual reading would be to do the products before the union operator.2012-09-14
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    "Am I suppose to impose some sort of topology on $X$ and $Y$...?" From the context, I'd say you are supposed to use the subspace topology, that is, the topology they inherit from the (usual) topology on the reals.2012-09-14
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    @tomasz my apologues for the typo on the first question. Y should be (X-{2}) union {1}. if there are continuous bijection between. X and Y. how do I get open sets from singletons, when there are open intervals?2012-09-14
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    @rick decker and Gerry Myerson I am not certain how to get open sets from the domain to show continuity. I have unit interval and sets with two elements even with the subspace topology.2012-09-14
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    For the second problem, the statement should read Construct the homeomorphism f:[0,1]×[0,1]→[0,1]×[0,1] such that f maps [0,1]×[0,1]∪{0}×[0,1] onto {0}×[0,1]. Also, f is a map means that f is continuous. Apologies for the misunderstanding.2012-09-14
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    Are you sure your latest wording is correct? $\{0\}\times[0,1]$ is a subset of $[0,1]\times[0,1]$ so $[0,1]\times[0,1]\cup \{0\}\times[0,1]$ should simply be $[0,1]\times[0,1]$, no matter how you interpret the order of operations.2012-09-14
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    @Rick, I think there is a typo at my last comment2012-09-14
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    @Rick, the map f should have read: f maps [0,1]×{0,1}∪{0}×[0,1] onto {0}×[0,1]. ([0,1]U{0})x({0,1}U[0,1]) And, [0,1]×{0,1}∪{0}×[0,1] is a subset of ([0,1]U{0})x({0,1}U[0,1]) which implies ([0,1]U{0})x({0,1}U[0,1]) is a subset of [0,1]x[0,1]. But what does [0,1]×{0,1}∪{0}×[0,1] look like? I am asked to create some kind of surjective function from the question.2012-09-14
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    @Rick I forgot to thank you in helping to format my original post properly and also pointing out the help section on the markup language. I am new to LaTex, but i will try to make my post here clear to read by others.2012-09-14
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    $[0,1]\times\{{0,1\}}$ is the top and bottom edges of a square. $\{{0\}}\times[0,1]$ is the left edge of that square. So the union is three of the four edges of the square. You're being asked to find a homeomorpshism of the square to itself that maps three edges of a square to one edge.2012-09-15
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    @Brian M. Scott Would the following work. let the codomain (0,1)=(0,1/2] union (1/2, 1) Then we define h piecewise as follows: h(y)=(1/2) * x if x is in (0,1] and h(y)=(-1/2)*(x-2)+1/2 Also, i am having a hard time tackling the second problem. As Gerry suggests, I have to map 3 edges to one edge. Can i split the one edge into 3 equal 1/3 portions, and from the domain, I map each edge to each portion? Thank you in advance2012-09-19
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    @GerryMyerson Gerry after rereading what you wrote, do i have to find a hoemomorphism of the square to itself, and also find a different homeomorphism that maps three edges of a square to one edge? Or am i to find only one homeomorphism. To find mapping of 3 edges to one edges, that is pretty easy. the one edge that gets map to, we just split it up to 3 pieces, and from the domain, each of the edge gets map to one of the pieces, and for the top and left edge, i use the pasting lemma for continuous functions, and do the same for the bottom and left edge.2012-09-21
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    Just one homeomorphism, a homeomorphism $f$ from the square to itself such that $f$ takes the three edges to the one edge.2012-09-21
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    @Gerry Myerson. so I just need to map 3 edges to one edge homeomorpically.2012-09-21
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    **NO**. When I write, "the square," I'm not just referring to the edges of the square; I mean the entire square, both edges and interior. Look at what the question says: $f:[0,1]\times[0,1]\to[0,1]\times[0,1]$. The homeomorphism has as its domain (and range) the entire square.2012-09-21

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