An open set of $ \mathbb{R}^2$ without a point is not simply connected
I need a rigorous proof of this because I only have the intuitive idea that a loop around the point can not be deformed into a point.
[Math] An open set of $ \mathbb{R}^2$ without a point is not simply connected
algebraic-topology
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Best Answer
Let $U$ be an open set and $x\in U$. Assume without loss of generality that $x=0$, the origin. We can define the continuous function
$$f:U\setminus\{0\}\to S^1, x\mapsto \frac x{\|x\|}.$$
Because $U$ is open, there is some $\epsilon>0$ so that $B(0,\epsilon)\subseteq U$. So take any closed curve $\gamma:[0,1]\to S^1$. You can embed this curve in $U\setminus\{0\}$ via $\gamma\mapsto \epsilon\gamma=:\gamma'$. If $U\setminus\{0\}$ is simply connected, there is a homotopy contracting $\gamma'$ to a point. Via $f$, this maps to a homotpy of $\gamma$ to a point in $S^1$. We know that $S^1$ is not simply connected, hence not every curve should be contractible. Contradiction. $\square$