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
Best Answer
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.