Prove that $$S=\{(x_1,x_2,...,x_{n+1})\in \mathbb{R}^{n+1}:x_1^{2}+x_2^{2}+...+x_{n+1}^2=1\}$$ is connected in $\mathbb{R}^{n+1}$
Definition:- Let $U \subset \mathbb{R}^{n+1}$. If for every pair of points $p$ & $q$ on $U$ , there is continuous function $\alpha:[a,b] \to U$ such that $\alpha (a)=p$ and $\alpha (b)=q$ then $U$ is called connected set in $\mathbb{R}^{n+1}$.
Please help me to prove this.
Let $p$ and $q$ be arbitrary points on $S$. Write $p=(p_1,p_2,\ldots, p_n)$ and $q=(q_1,q_2,\ldots, q_n)$. So $\sum_{i=1}^n p^2_i = \sum_{i=1}^n q^2_i = 1$.
Let $\alpha: [0,1] \mapsto \mathbb{R}^n$ be the following function:
$\alpha(t) = y(t)$, where $y(t)$ is defined $y_i(t) = \sqrt{tq^2_i + (1-t)p^2_i}$ for each $i=1,\ldots, n$. As $\sum_i y^2_i(t)$ is 1 for all such $t \in [0,1]$, it follows that $\alpha$ is indeed a function from $[0,1]$ to $S$. Furthermore, $\alpha$ is clearly continuous. Finally, $\alpha(0) = p$ while $\alpha(1) = q$.
The accepted answer given by Mike looks great, but upon closer examination you have to ask how to define $y_i(t)$ when, say, $p_i > 0$ and $q_i < 0$.
Here is a simple proof that, like Mike's idea, only uses the square root function; it just applies the technique found here
Prove that the unit circle is path-connected
Sketch:
We can use the mappings
$\tag 1 t \mapsto \left(\,(1-t)x_0,\, \pm\sqrt{r^2 - [(1-t)x_0]^2}\,\right) \text{ with } 0 \lt r \le 1$
to continuously connect (push), in turn, all the $x_k$ coordinates, for $1 \le k \le n$, to $0$. After this you are positioned at either $(0,0,\dots,-1)$ or $(0,0,\dots,+1)$. But using
$\tag 2 t \mapsto \left(\,\sqrt{1-t^2},\, t\,\right)$
we can ultimately connect any point on the $n\text{-sphere}$ of $\mathbb R^{n+1}$ to $(0,0,\dots,+1)$.
Yes @CopyPasteIt you are right. Implicitly assumed in my argument is that $p_iq_i$ is nonnegative for each $i$. And this clearly is not the case. We can get around this though: Let $S^+$ be the set of $i$ s.t. both $p_i$ and $q_i$ $\ge 0$ and let $S^-$ be the set of $i$ s.t. both $p_i$ and $q_i$ $\le 0$. Two cases:
Case 1: $|S^+ \cup S^-| = a \ge 1$. Then let $m$ be the point s.t. $m_i = \frac{1}{\sqrt{a}}$ if $i \in S^+$; $m_i = -\frac{1}{\sqrt{a}}$ if $i \in S^-$, and 0 otherwise. Use my line of reasoning in my previous Answer about to conclude that $p$ is connected to $m$ and $m$ is connected to $q$.
Case 2: $|S^+ \cup S^-|$ is empty. Then let $i_1$ and $i_2$ be two distinct integers. Let $m$ be the vector such that $m_{i_1} = $sign$(q_{i_1})(\sqrt{2})^{-1}$ and $m_{i_2} = $sign$(p_{i_2})(\sqrt{2})^{-1}$ and $m_i = 0$ for every other $i$. Use the line of reasoning in Case 1 to conclude there is a path from $q$ to $m$, and then from $p$ to $m$.
