Topology: Determine whether a subset is a retract of R^2

[nightingale123]

Homework Statement

Let $X=([1,\infty)\times\{0\})\cup(\cup_{n=1}^{\infty}\{n\}\times[0,1])$ and $Y=((0,\infty)\times\{0\})\cup(\cup_{n=1}^{\infty}\{n\}\times[0,1])$

$a)$Find subspaces of of the euclidean plane $\mathbb{R}^2$ which are homeomorphic to the compactification with one point $X^+$ and $Y^+$

$b)$ are any of the subspaces $X$ and $Y$ retracts of $\mathbb{R}^2$

Homework Equations

The Attempt at a Solution

I was able to find the required subspaces for $a)$ and proved that they are homeomorphic, however I'm having some problems with $b)$. I know that $Y$ cannot be a retract as it is not closed.

I got that since $\mathbb{R}^2$ is a metric space therefore $T_{2}$ and we know that if $f,g:X\to Y$ are continuous and $Y\in T_{2}$ then the set $\{x|f(x)=g(x)\}\subset Y$ is closed. However in this case $Y=\mathbb{R}^2$ and if there exists a retraction $r:\mathbb{R}^2\to Y\subset \mathbb{R^2}$ the set $\{x|r(x)=id_{x}(x)=x\}$,which is $X$, must be closed. And since it is not. $X$ cannot be a retraction. Can someone correct me here if I did something wrong?

However this argument does not work for $X$ as $X$ is closed. $X$ is also connected so that fails too. I can't find a homeomorphism between $X$ and $\mathbb{R}$. Could someone clarify how we could determine whether X is a retract or not?

EDIT: I'm sorry for posting this in the advanced physics section. I just noticed it and I don't know how to change it

 

[andrewkirk]

I have an intuition about X that may or may not turn out to be a hint that leads to a proof that X is not a retract. Call the retract function f.

X is a 'comb' that is an infinite horizontal row of vertical segments (tines) stuck onto the comb spine that is the x-axis to the right of (1,0).

Let B be X with the spine removed. If there exists a tine that does not intersect f(~X) then we'd expect there to be some sort of discontinuity at the tip of that tine, because points in ~X infinitesimally close to the tip must map to the spine while the tine tip must map to itself.

On the other hand, if every tine intersects with f(~X), we could partition ~X into pre-images of the tines and a pre-image of the spine, and we might expect to run into discontinuity problems at the boundaries of these pre-images.

 

[nightingale123]

I'm sorry but you kinda lost me at this part

Let B be X with the spine removed. If there exists a tine that does not intersect f(~X) then we'd expect there to be some sort of discontinuity at the tip of that tine, because points in ~X infinitesimally close to the tip must map to the spine while the tine tip must map to itself.

On the other hand, if every tine intersects with f(~X), we could partition ~X into pre-images of the tines and a pre-image of the spine, and we might expect to run into discontinuity problems at the boundaries of these pre-images.

 

[andrewkirk]

I'm sorry but you kinda lost me at this part

For the first para, say for argument's sake the first tine $T_1\equiv \{1\}\times (0,1]$ does not intersect f(~X). Consider the tine's tip P=(1,1). We know $f(P)=P=(1,1)$. But for any point $Q\notin X$ that is near P, f(Q) is not in $T_1$ so it will either be somewhere on the x-axis or on a different tine. In either case f(Q) will be distance at least 1 away from f(P), regardless of how close Q is to P. Can you use that to prove that f is discontinuous at P, which would mean that f is not a retraction function?