I understand why $R^3 – {(0,0,0)}$ is simply connected, and I also understand why $R^2 – {(0,0)}$ is not simply connected. The way I look it at is if checking if the region is $a)$ path-connected and $b)$ any curve can be contracted to a point in the region.
From what I reasoned it seems there is a pattern, a hole in $R^2$ prevents simply connectedness, a missing line in $R^3$ does the same, and so I reasoned for $R^n$ any $n-2$ dimensional missing figure (or higher) would prevent the region from being simply connected.
Then I was posed with the scenario: take $D$ to be all of $3D$ space except for a sphere of radius 1, is $D$ simply connected? The answer is apparently yes, D is simply connected because "the spherical hole does not prevent paths from contracting to points while remaining in $D$". However, now I'm confused because a spherical hole is a $3D$ hole and it goes against my previous conjecture. Also, according to this MIT video: https://www.youtube.com/watch?v=6S3BJSsc72Q , $R^3 -$ a circle is not simply connected.
So why is $R^3$ – a spherical hole simply connected?
Best Answer
Think about loops with some fixed point in $\mathbb{R}^3$. You need to be able to contract $\it{any}$ possible loop attached to that fixed point back to the fixed point. For the spherical hole (deleted ball) you can always pass the loop around the ball. Now picture a finite length pole with a loop around it. You can still pass the loop around the rod, but not through it. If the rod were infinite in length (a line in $\mathbb{R}^3$) you could neither pull the loop through the rod or around the end of the rod.