$X=\{(x,y)\mid x = r\cos(\frac{\pi}{2^n}), y = r\sin(\frac{\pi}{2^n}), 0\leq r\leq1, n = 2,3,4, \ldots\}$ $\cup \{(0,1)\}$ with the usual topology.

Question

Well they give the following subset:

$X=\{(x,y)\mid x = r\cos(\frac{\pi}{2^n}), y = r\sin(\frac{\pi}{2^n}), 0\leq r\leq1, n = 2,3,4, \ldots\} \cup \{(0,1)\}$ with the usual topology.

And they ask me:

1. Is compact?
2. Is connected? If it isn't connected find it components.
3. Is path connected? If it isn't path connected find it path components.

Well the first thing I have done is draw $X$ in $\mathbb{R^2}$ and also see that the cos and sen are between $[\frac{\pi}{4},0)$ then I think that is isn't compact because if we take $(r\cos(\frac{\pi}{2^n})-\epsilon,r\cos(\frac{\pi}{2^n})+\epsilon)$x$(r\sin(\frac{\pi}{2^n})-\epsilon,r\sin(\frac{\pi}{2^n})+\epsilon)$ that is an infinite cover for $X$ we can't obtain any finite subcover for it. For the connected and path connected I don't know how to prove that they are connected and path connected.

Answer 1

1. $X$ is not compact because $X$ is not closed. Indeed $X^c$ is not open because for any $\epsilon >0$ the disk $D((1,0), \epsilon)$ has a non-empty intersection with $X$. However $X$ is relatively compact since it is bounded in $\mathbb{R}^2$.
2. and 3. $X$ is disconnected : in the induced topology of $X$, $\{(0,1)\}$ is open and disjoint to $X' = X \setminus \{(0,1)\}$ which is also open. However $X'$ is path-connected because it is star-shaped, which means there exists a single point $x_0$ (in this case $(0,0)$), such that for any $x\in X'$, $[x_0,x] \subset X'$. Therefore $X'$ is connected.

So the two connected components of $X$ are $\{(0,1)\}$ and $X'$ wether we talk about connectedness or path-connectedness.