Is $P$ is homeomorphics to $Q$? yes/NO

Question

Is $P$ is homeomorphics to $Q$?
$P=\{(x,y)\in \mathbb{R^2}:y=\sin (\frac{1}{x}),0<x\le 1\}\cup \{(x,y)\in \mathbb{R^2}:,x=0,-1\le y\le 1\}$

$Q=[0,1] \subseteq \mathbb{R}$.

My attempt : i know that P is topologist sine curve it is compact and connected and $[0,1] $ is also compact and connected by heine borel theorem

i thinks both P and Q are homoeomorphics

is my logics is correct/not correct?

any hints/solution will be appreciated

thanks u

Answer 1

As an easy way to see that your reasoning is not correct, note that any subset of $\mathbb{R}$ containing exactly one point is compact and connected, but is not homeomorphic to the interval $[0,1]$.

As regards $P,Q$, note that if you remove the $3$ points $$(0,1),\;\;(0,-1),\;\;(1,\sin(1))$$ from $P$, the remaining set $S$ is still connected.

But if you remove any $3$ distinct points from $[0,1]$, the remaining set is disconnected.

If $f:P\to Q$ was a homeomorphism, the restriction of $f$ to $S$ would be a homeomorphism from $S$ to $f(S)$. But then, since $f(S)$ is equal to $[0,1]$ with $3$ points missing, $f(S)$ is not connected, contradiction.

It follows that $P,Q$ are not homeomorphic.

Here's another way to see it . . .

Since the removal of any of the $3$ points $$(0,1),\;\;(0,-1),\;\;(1,\sin(1))$$ doesn't disconnect $P$, a homeomorphism $f:P\to Q$ would have to map each of those $3$ points to either $0$ or $1$, but then the pigeonhole principle would imply that $f$ is not injective, contradiction.

Answer 2

Your logic is not correct. You do not have enough to assert that your $P$ and $Q$ are homeomorphic. You are essentially guessing (I do not mean this in a bad way).

HINT:

Path-connectedness is invariant under homeomorphism. That is to say, if a path connected space $X$ is homeomorphic to $Y$, then $Y$ has to be path-connected too.