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Let $A$ be the following subset of $\Bbb R^2$:

$$A=\left\{(x,y)\in\Bbb R^2:(x+1)^2+y^2\leq 1\right\}\cup \left\{(x,y)\in\Bbb R^2: y=x\text{sin}\left(\frac{1}{x}\right),x>0\right\}$$.

Prove or disprove the set is path connected.

Definition: A space $X_\tau$ is path connected if for every pair of points $x_0,x_1\in X$, there exists a path $\alpha:[0,1]_\mathfrak{U}\to X$ with $\alpha(0)=x_0$ and $\alpha(1)=x_1$.

How to solve above problem?

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  • $\begingroup$ I'm a little confused about the definition of $A$. It is the union of two sets, the first of which is presumably the unit disk centred at $(-1, 0)$ (you should use set-builder notation here!), while the other seems like a perfectly sensible subset of $\Bbb{R}$. Are you embedding $\Bbb{R}$ into $\Bbb{R}^2$ the usual way? Is the second set supposed to be the graph of a function? Or are you endowing the set $\Bbb{R}^2 \cup \Bbb{R}$ with some funky topology (and if so, what is it)? $\endgroup$
    – user837206
    Oct 25, 2020 at 5:51
  • $\begingroup$ Sorry , I edited. Now see $\endgroup$
    – user787636
    Oct 25, 2020 at 6:01
  • $\begingroup$ @Joseph: In connection with the answer by Cronus you may find my answer here helpful. $\endgroup$ Oct 25, 2020 at 17:41

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First use the fact that a union of two path-connected subsets, whose intersection is nonempty, is path-connected. Then try to prove for each of them separately that they're path-connected.

The set on the left is a ball, so it's convex. That is, it's pretty easy to see that the straight line between any two points inside it is contained in it, and hence is a path between them.

The set on the right is the graph of a continuous function (well, you need to prove that it is continuous, I guess). Call this function $f$. So any two points in it are of the form $(x_1,f(x_1))$ and $(x_2,f(x_2))$, and it's easy to see the path $\varphi:[x_1,x_2]\to\Bbb{R}^2$ defined by $\varphi(t)=(t,f(t))$ is a path between them.

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  • $\begingroup$ As given $x>0$ Is it possible they have a point in common? $\endgroup$
    – user787636
    Oct 25, 2020 at 11:37
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    $\begingroup$ @Joseph Yeah you're right, I was being sloppy when I said "the right hand side". You first need to observe that $A=B_1(-1,0)\cup \{(x,f(x)):\geq 0$ (where $B_1(-1,0))$ is the ball of radius $1$ around $(-1,0)$), since $(0,f(0))=(0,0)$ is contained in the ball. Then you proceed as I said. $\endgroup$
    – Cronus
    Oct 25, 2020 at 16:21
  • $\begingroup$ Thank you ${}{}{}$.! $\endgroup$
    – user787636
    Oct 26, 2020 at 2:05

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