Prove this set is not connected

Question

I have this set

$A=\{(0,0,1),(0,0,-1)\}\subset \mathbb R^3$

How can I prove this is not path connected (or not connected)?

Thanks.

Answer 1

It's not connected by definition. ($A$ is a union of two disjoint nonempty open sets)

Path connected implies connected, so in the same way if it's not connected it's not path connected. And you are done!

EDIT you may think $\{(0,0,1)\}$ is a closed subset of $A$ - which it is. But by the subspace topology this means that $A\setminus\{(0,0,1)\} = \{(0,0,-1)\}$ is open. Similarly you can show that $\{(0,0,1)\}$ is open (so both of these subsets are clopen disjoint nonempty subsets).

Answer 2

$A$ can be expressed as the disjoint union of two open sets $U$ and $V$.

let $a=\{(0,0,1)\}$ and $b=\{(0,0,-1)\}$ Define $U=B_{\epsilon}(a)$ and $V=B_{\delta}(b)$ (open balls with radius $\epsilon$)

Now the trick is to find an $\epsilon$ and $\delta$ small enough that $U$ and $V$ are disjoint. Pick $\epsilon=1$ and recall the radius is $\epsilon=(x^2+y^2+z^2)^{1/2}$, and around $(0,0,1)$ we have $B_{1/2}(a)=\{(x,y,z):((x-0)^2+(y-0)^2+(z-1)^2)^{1/2}<1)\}$

Note that the point $\{(0,0,1/n)\}$ is contained in $U$ for every $n\in \mathbb{N}$. Now take $V$ to be the ball with radius $\delta=1$, and make a similar argument, this time for $-1/n$. The origin is contained in the union, and $U$ and $V$ are disjoint, with $a\in U$ and $b\in V$, so the set $A$ is not connected, and by contrapositive, cannot be path-connected. You also could have toyed around with $\epsilon$ and $\delta$ but setting them both to $1$ makes the argument neater.