How to prove a space is path-connected

Question

If we consider the space given by $\{(x_{1}, x_{2}, ..., x_{n}) : (x_{1}, x_{2}) \neq (0, 0)\} \subset \mathbb{R}^{n}$, for $n \ge 3$, how can we show that it is path-connected?

For $n=3$, the space is just $\mathbb{R}^{3}$ with the z-axis removed, so intuitively I know to construct a straight line composition around the removed axis, but not sure how to put do it explicitly.

As for higher dimensions I am really stuck as I cant even visualize what the space looks like. Was thinking maybe a homeomorphism from a space we know is path connected such as $\mathbb{R}^{n}-\{0\}$, since it would preserve the path-connectedness condition.

Answer 1

A down-to-earth approach would be the following.

Every point $(x_0,...,x_n)\in X=\{(x_0,...,x_n)\mid (x_0,x_1)\neq 0\}$ is connected to some point in $X'=\{(x_0,x_1,0,...,0)\mid (x_0,x_1)\neq 0\}$ by letting $$\begin{array}{rcl} [0,1]&\rightarrow&X\\ t & \mapsto & (x_0,x_1,(1-t)x_2,...,(1-t)x_n) \end{array}$$ Hence it suffices to show that $X' \cong \Bbb R^2\setminus \{0\}$ is pathconnected.

For this we note that every point in $v\in\Bbb R^2 \setminus \{0\}$ is connected to some point in $\Bbb S^1$ by letting $$\begin{array}{rcl} [0,1]&\rightarrow &\Bbb R^2\setminus \{0\}\\ t & \mapsto &t \dfrac{v}{\Vert v\Vert} + (1-t)v \end{array}$$

Finally $\Bbb S^1$ is pathconnected.

Another approach (or rather the generalization of the proof above) is the one suggested by Andrew D. Hwang. We have an isomorphism $X \cong (\Bbb R^2\setminus \{0\}) \times \Bbb R^{n-1}$ and a product of two spaces is pathconnected if and only if both of them are. Hence it suffices to show that $\Bbb R^2 \setminus \{0\}$ is pathconnected. The immediate generalization of the argument above shows that a space, which admits a pathconnected deformation retract (in our case $\Bbb S^1$), is pathconnected as well.

Answer 2

A constructive approach:

Take two points $p_1$ and $p_2$. Let $l$ be the line $l(t)=tp_1 + (1-t)p_2$ for $t\in [0,1]$. Suppose $l$ is such that $l(t_0)=(0,0,x_3,\ldots, x_n)$ for some $t_0\in (0,1)$. We now look at three cases:

Case 1: Suppose the first coordinate of $p_1$ and $p_2$ coincide and are both zero. Then the second coordinates of $p_1$ and $p_2$ are both non-zero (else $p_1$ or $p_2$ would have first two coordinates zero), and also do not coincide (else the line $l(t)$ would never have second coordinate zero). So then $t_0$ is the only value where $l(t)$ has second coordinate zero. Then the line $$ l'(t) = l(t) + (\sin(\pi t),0,\ldots, 0) $$ is as desired. To see this, note that $l(t)$ has second coordinate zero only when $t=t_0$. But then the first coordinate is $0 + \sin(\pi t_0)\neq 0$.

Case 2: The first coordinate of $p_1$ and $p_2$ coincide but are not zero. Then $l(t)$ has first coordinate non zero for all $t$. So this cannot happen as we assume $l(t_0) = (0,0,x_3,\ldots, x_n)$.

Case 3: The first coordinates of $p_1$ and $p_2$ do not coincide. Then $t_0$ is the only value where the first coordinate of $l(t)$ is zero. Then the line $$ l''(t) = l(t) + (0,\sin(\pi t),0,\ldots, 0) $$ is as desired.