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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

1 Answers 1

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For the first question, you have $$X=\bigcup_{n\in\Bbb N}(3n,3n+1)\cup\{3n+2:n\in\Bbb N\}$$ and $Y=\big(X\setminus\{2\}\big)\cup\{1\}$. Thus, $$X=(0,1)\cup\{2\}\cup(3,4)\cup\{5\}\cup(6,7)\cup\{8\}\cup\ldots\;,$$ and $$Y=(0,1]\cup(3,4)\cup\{5\}\cup(6,7)\cup\{8\}\cup\ldots\;.$$ You are to consider each of these spaces with the topology that it inherits from the usual topology of $\Bbb R$. Your bijection

$$f:X\to Y:x\mapsto\begin{cases}1,&\text{if }x=2\\ x,&\text{if }x\ne 2 \end{cases}$$

is in fact continuous with respect to these topologies, and you shouldn’t find this too hard to show: the only point of $x$ at which it could possibly not be continuous is $2$, and to show that $f$ is continuous at $2$, just use whatever definition of continuity you have available.

Getting a continuous bijection $g:Y\to X$ is a bit harder. Here’s a hint: find a continuous bijection $h:(0,1]\cup(3,4)\to(0,1)$, and then define

$$g:Y\to X:y\mapsto\begin{cases} h(y),&\text{if }y\in(0,1]\cup(3,4)\\ y-3,&\text{otherwise}\;. \end{cases}$$

Don’t try to be fancy with $h$: the simplest possible idea works.

Finally, you’ll have to prove that $X$ and $Y$ aren’t homeomorphic. The key is the point $1\in Y$. Suppose that you have a homeomorphism $h:Y\to X$, and ask yourself where $h(1)$ can be. Show first that it can’t be one of the isolated points of $X$ (e.g., $2,5,8$). Then show that it can’t be in any of the open intervals $(3n,3n+1)$, either; for this you’ll want to use connectedness. For now I’ll not say just how, but if you get stuck, please ask.

Gerry Myerson has explained in the comments what is being asked in the second question, so I’ll leave that alone unless you have further questions.