Does a compact path-connected set always have a finite number of holes?

Question

Does a compact path-connected set always have a finite number of holes?

I'm visualizing something like this:

https://en.wikipedia.org/wiki/Simply_connected_space#/media/File:Runge_theorem.svg

Also, if I'm in euclidean space, can I further say that each of these holes has a finite area?

Answer 1

How about this. Let $C$ be the usual Cantor set. Consider $X=(C\times[-1,1])\cup ([0,1]\times\{-1,1\})$ in the plane. So put a vertical segment through each point of $C$ and close them up at the top and bottom. A path-connected compact space inside $\Bbb R^2$, which I hope you agree has infinitely many holes.

Answer 2

In $\mathbb R^2$, take the closed ball of radius $1$ around the origin, and remove an open ball of radius $1/n^3$ around $(0, 1/n)$ for $n=2,3, 4,\ldots$.

What remains is a closed and bounded set, and therefore compact. It's easily seen to be path-connected. And there are infinitely many holes because the balls you removed do not overlap.