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Is it possible for the closure of a simply connected domain in the complex plane to not be simply connected? Intuitively it seems the closure is simply connected but I can't prove it.

Is it enough to show that every point is homotopic to some point in the interior?

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up vote 9 down vote accepted

The set $X = \{re^{i\theta}\ | 1 < r < 2, -\pi < \theta < \pi\}$ (an open annulus without the negative real axis), is simply connected but $\overline{X} = \{re^{i\theta}\ | 1 \leq r \leq 2\}$ is not simply connected.

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Consider an annulus with a ray removed: $$\{re^{i\theta}: r\in(1,2), \theta\in(0,2\pi)\}$$

This is the homeomorphic image of a rectangle, hence simply connected. However, its closure is the annulus $\{z : |z|\in[1,2]\}$, which is not simply connected.

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