I had a thought about the path-connectedness of $\mathbb{R}$. The standard way of showing the path-connectedness of $\mathbb{R}$ can be that it is both open and connected and so must be path-connected too.
But I had thought about another argument, say $a$, $b\in\mathbb{R}$, can we just define a path that connects $a$ and $b$ as follows? $\gamma:[0, 1]\to \mathbb{R}$ such that $\gamma(t)=a+(b-a)t$ which precisely take $a$ to $b$ and the image precisely lie in $\mathbb{R}$ as well since $\mathbb{R}$ is closed under multiplication and addition?
Obviously, the first method sounds much more rigorous but it will be much longer to prove so I was wondering if the second method is insufficient in proving the path-connectedness?
Best Answer
First one shows that in a linearly ordered space topological space $X$ (as $\Bbb R$ is), $X$ is connected iff it has no gaps and no jumps. As $\Bbb R$ has a dense order, it has no jumps and as it has the lub-property, it has no gaps. It follows that $\Bbb R$ and $[0,1]$ are both connected.
That $\Bbb R$ is then path-connected follows from your second argument: we can find a path between any two points by using the continuous addition and multiplication, and this argument works in any $\Bbb R^n$ or topological vector space over $\Bbb R$. Paths are defined on subsets of $\Bbb R$ and so that works really well. Path-connectedness implies connectedness just because we know that $[0,1]$ is already connected. So connectedness of intervals is the most basic fact upon which we rely.