Proving that $S^n$ (n-sphere) is locally connected.

Definition: A space X is said to be locally connected at $x\in X$ if for any open set $U$ containing $x$, there is an open connected subset of $U$ (say $W$) containing $x$.$$x\in W\subseteq U$$ A space X is said to be locally connected if it is locally connected at each $x\in X$

Equivalent statement: A space $X$ is locally connected iff its basis consists of open connected sets.

I tried proving the claim by using the second statement.

$\mathfrak{B}=\{S^n\cap B_\epsilon(x)\,\,|\,\,x\in S^n\}$ would form a basis for $S^n$, where $B_\epsilon(x)$ is a ball of radius $\epsilon$ centered at $x$.

Now, I tried proving $U=S^n\cap B_\epsilon (x)$ is open (which holds due to subspace topology induced from $\mathbb{R}^{n+1}$) and connected.

I am having problem showing that $U$ is connected. One approach would be to show $U$ is path connected (which would imply connectedness). I know we can show path connectedness by showing that each of the points can be path connected to a common point $x_0$

Let $y\ne x_0$, then will the following map work? $$f:[0,1]\to U$$ $$f(t)=\frac{ty+(1-t)x_0}{||ty+(1-t)x_0||}$$

I don't know whether the image of $f$ will lie entirely in the domain or not.

I would also like to know if the map gives "a" shortest path between $x_0$ and $y$

Any help would be appreciated. Thanks!

• Hint: your map $f$ is well-defined and will work unless $ty+(1-t)x_0 = 0.$ When will this happen? – jflipp Apr 20 '15 at 10:41
• And yes, if the map $f$ is well.defined, it is a shortest path between $x_0$ and $y.$ In fact, $f$ is an arc on the so-called great circle through $x_0$ and $y.$ This great circle is the intersection of $S^n\subseteq \mathbb R^{n+1}$ and the plane spanned by $x_0,y,$ and the origin. $f$ is a shortest path because the great circles are the geodesics on $S^n,$ see here. – jflipp Apr 20 '15 at 10:45
• @jflipp when the points are diametrically opposite? – Swapnil Tripathi Apr 20 '15 at 11:02
• I think you need to show that $||f(t)-x_0||<\epsilon$. – Stefan Hamcke Apr 20 '15 at 11:42