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Why is $\mathbb{R^2} \setminus (0, 0)$ connected but not simply connected?

Simply connected means path connected, intuitively it seems that for any point in $\mathbb{R^2} \setminus (0, 0)$ we can have an unbroken path to any other point in this set? So what am I misunderstanding?

Michael Hardy
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sonicboom
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    Simply connected means that each closed path is homotopy-equivalent to a constant map. But the closed path $x\mapsto \langle\cos x, \sin x\rangle$ is not. – MJD Apr 01 '14 at 17:03

3 Answers3

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Simply connected does not mean path-connected.

Simply connected means that if a path starts at a point $p$ and returns in to $p$, then that loop can be contracted to $p$.

If a path starts at $1$ and winds once clockwise around $0$ and returns to $1$, then that path cannot be contracted to $1$ within the space $\mathbb R^2\setminus \{(0,0)\}$. It gets "caught" on the point $(0,0)$.

Michael Hardy
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  • Just happened to pass by..(because I have no experience in dealing with things like this).. it seems that removing any single ordered pair from $\Bbb R^2$ would violate the condition for simply connected. But is there any other way it can be not simply connected? I ask because I think a new term wouldn't have been created if the need wasn't broader. – Guy Apr 01 '14 at 17:30
  • There are many other ways. You can, for example, show that the differential form $\frac{ydx}{x^2 + y^2} - \frac{xdy}{x^2 + y^2}$ is closed, but not exact. With a detour through cohomology, this will show that it is not simply connected. – Simon Rose Apr 01 '14 at 17:39
  • @Sabyasachi : You can remove many isolated points, or remove whole disks or sets of many other kinds, and get a space that is not simply connected. The torus is also not simply connected. The projective plane is not simply connected. The ordinary circle is not simply connected. There are many manifolds that are not simply connected. – Michael Hardy Apr 01 '14 at 20:54
  • @MichaelHardy ah I see now. Thanks. – Guy Apr 02 '14 at 04:26
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Your definition is incorrect: simply connected means that any loop in the space can be continuously shrunk to a point. But a loop around the missing point of $\mathbb R^2-\{(0,0)\}$ (for instance, a parameterization of the unit circle centered at the origin) cannot be shrunk to a point in a continuous manner without going through the missing point $(0,0)$; there are formal proofs of this, but I think it's obvious enough to "see".

user134824
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Path connectedness is a more basic concept. Simple connectedness means any two paths between two points can be 'deformed continuously' to each other. Take the unit circle centred at origin. The upper and lower semi-cricular arcs can be thought of as two paths from $(-1,0)$ to $(1,0)$. When we try to push one of these arcs towards the other you have a hurdle at (0,0). SOme intermediate path has to use the origin which unfortunately has been removed.

P Vanchinathan
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