Is this set path-connected?

Question

Consider $\left\{(e^{-x}\cos x, e^{-x}\sin x)\in \mathbb{R}^2: x\geq 0 \right\} \cup \left\{(0,0) \right\}.$ I need to see if this set is path connected. My guess is it is not primarily Because of the point $(0,0)$. There needs to be a path from $(0,0)$ to any other point in the set. Since the other points are of the form $(e^{-\alpha}\cos \alpha, e^{-\alpha}\sin \alpha)$ for some $\alpha\geq 0,$ I must have that there is some $\alpha$ that will put the point $(e^{-\alpha}\cos \alpha, e^{-\alpha}\sin \alpha)$ arbitrarily close to $(0,0)$. Since, $\cos x$ and $\sin x$ are never zero at the same time we need to send $\alpha$ to $\infty.$ What we seem to have is the closure of the set $$\left\{(e^{-x}\cos x, e^{-x}\sin x)\in \mathbb{R}^2: x\geq 0 \right\}.$$ Am I going in the right direction? How would I prove that there is no path from $(0,0)$ to any other point in $\left\{(e^{-x}\cos x, e^{-x}\sin x)\in \mathbb{R}^2: x\geq 0 \right\}$?

Answer 1

Let $\gamma$ be the path $$ \gamma(t) = \begin{cases} (0, 0) &\text{ if } t = 0 \\ (e^{-\frac 1 t}\cos(\frac 1 t), e^{-\frac 1 t}\sin(\frac 1 t)) & t \in (0, 1] \end{cases} $$ This is continuous. To show this, it's clear that $\gamma$ is continuous on $(0, 1]$ so we only need to consider $t = 0$. Note that $$ |(e^{-\frac 1 t} \cos(\frac 1 t), e^{- \frac 1 t}\sin(\frac 1 t))| = e^{-\frac 1 t} $$ And so $\gamma$ is continuous at 0 using the $\varepsilon$, $\delta$ definition of continuity.

As you say, the set is path connected if any two points can be connected by a path. It's clear that for all $x \geq 0$ we can connect $(e^{-1}\cos(1), e^{-1}\sin(1))$ to $(e^{-x}\cos(x), e^{-x}\sin(x))$, and so concatenation with $\gamma$ will give a path connecting $(0, 0)$ to any point. Hence the set is path connected.