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It seems this is too obvious to be written anywhere, but how do you prove that the 3-sphere is connected?

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Suspensions of nonempty spaces are always path connected, and $S^3 = S(S^2)$. (Overkill, I know.) –  Thomas Belulovich Apr 25 '12 at 16:00

3 Answers 3

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If $x$ and $y$ are points on $S^3$, connect them by a path, e.g., if $P$ is a plane which contains $0$, $x$, and $y$ in $\mathbb{R}^4$, $P\cap S^3$ will be a circle in $S^3$ which contains both points. Taking one component of $P\cap S^3 - {x,y}$ will give a path between $x$ and $y$.

Now recall that path-connected implies connected and you're done.

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The $n$-sphere is path connected for $n \ge 1$. We can prove by induction on $n$ that every point has a path to $(0,\ldots,0,1)$:

Let the given point be $(x_1,x_2,\ldots,x_{n+1})$. If $x_1$ is not already $0$, then fix $x_3$ through $x_{n+1}$, and the valid choices of $x_1$ and $x_2$ are now the circle defined by $x_1^2+x_2^2 = 1-x_3^2-\cdots-x_{n+1}^2$. Go around this circle until you reach one of the points where $x_1=0$. Then unfix $x_3$ and proceed by induction in the $(n-1)$-sphere defined by $x_1=0$ (i.e. the equator of the original $n$-sphere).

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If you prefer to play with only connectedness (and avoid path-connectedness arguments), you may also consider the map $f:\mathbb{R}^{4}\setminus \{0\}\to S^{3}$ so that $f(x)=\frac{x}{||x||}$, which is a continuous surjection.

It is quite elementary that a continuous image of a connected set is connected, so infact you only need to show that $\mathbb{R}^{4}\setminus \{0\}$ is connected.

If you would have a separation of $\mathbb{R}^{4}\setminus \{0\}$, would you be able to derive a contradiction through the connectedness of $\mathbb{R}^{4}$?

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