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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?

Grid
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    Do you mean $3D$ space expect for a sphere or a ball ? – Stefan Hamcke Apr 09 '14 at 16:35
  • This is the question word for word: "Take $D$ to be all of three-dimensional space except for a sphere of radius $1$ centered at $(4,2)$. Is $D$ simply connected?" I'm not sure why the center has only two coordinates, but that's the question the instructor gave. – Grid Apr 09 '14 at 16:41
  • But a sphere disconnects the space into a bounded and an unbounded region, so the complement is not connected. He probably meant a ball. – Stefan Hamcke Apr 09 '14 at 16:43
  • This may sound like a naive question, but what is the difference between a "ball" and a "sphere"? – Grid Apr 09 '14 at 16:45
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    A sphere is just the surface, all points whose distance from the origin equals $1$. A ball is a filled sphere, all points whose distance is less (for the open ball) or $\le$ (for closed ball) than $1$. Note that in $\Bbb R^n$, the ball is denoted by $B^n$ (or $D^n$) while its boundary, the sphere, is denoted by $S^{n-1}$. – Stefan Hamcke Apr 09 '14 at 16:47

4 Answers4

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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.

Ryan Vitale
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    Thanks, that clears up a lot, but you can't do the same with a circle in 3-dimensional space? – Grid Apr 09 '14 at 16:55
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    No, because you can make a loop around the circle which you "can't pull through it". Think about two rings linked together (more like one rigid hoop for the circle, and a loop of string around the hoop that you'd like to contract down to a point) – Ryan Vitale Apr 09 '14 at 16:57
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$\mathbb{R}^3$ minus a 3-dimensional ball is simply connected.

$\mathbb{R}^3$ minus a sphere (e.g. minus the surface of a ball) is not, though for a slightly different reason than that $\mathbb{R}^3$ minus the $z$-Axis is not.

In $\mathbb{R}^3$ minus a sphere, your region consists of two disconnected parts - the inside of the sphere and the outside. You can continuously transform every path into a point, but you can not transform continuously transform every path into every other path - you can only do that if the paths reside on same side of the sphere.

fgp
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Note that $A:=\mathbb{R}^{3}\setminus S^{2}$ is not connected and thus not path connected, which breaks the definition of being simply connected already. So the answer isn't very deep.

In fact, to define $\pi_{1}(A)$, which is base-point free notation, you need $A$ to be path connected. So the notation $\pi_{1}(A)$ is already a little ambiguous. For non path-connected spaces you have to study each $\pi_{1}(A,x_{0})$ separately with $x_{0}$ being a base point at a given path component. Each of the path components of $A$ are simply connected, though.

T. Eskin
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Firstly, R^3 - {origin} is simply connected. The reason is that, naively, you have more degree of freedom as far as finding a continuous path in the region is concerned.

Yash
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