# homeomorphism related question, how to show two space is homeormophic to each other or not

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.

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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 $\{2\}$? – tomasz Sep 14 '12 at 1:48
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. @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. – Rick Decker Sep 14 '12 at 1:52
"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. – Gerry Myerson Sep 14 '12 at 3:13
@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? – Seth Mai Sep 14 '12 at 8:37
@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. – Seth Mai Sep 14 '12 at 8:39

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.

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