$\Gamma$ is a path connected

Solution 1:

For $x\in \Bbb R$ and $t\in [0,1]$ let $f_x(t)=(x,t)\in\Gamma.$

Let $B$ be the usual base (basis) for $\Gamma,$ i.e. open disks in $\Bbb R\times \Bbb R^+$ that are not tangent to $\Bbb R\times \{0\},$ along with $D\cup \{p\}$ when $D$ is an open disk in $\Bbb R\times \Bbb R^+$ that's tangent at $p$ to $\Bbb R\times \{0\}$.

Show that if $b\in B$ then $f_x^{-1}b,\;$ i.e. $f_x^{-1}(\,b\cap (\{x\}\times [0,1])\,),$ is open in the space $[0,1].$ So $f_x$ is continuous. So any point $(x,0)$ in $\Bbb R\times \{0\}$ is path-connected to a point $(x,1)$ in the path-connected subspace $\Bbb R\times \Bbb R^+.$ (Note that the topology on $\Bbb R\times \Bbb R^+$ as a subspace of $\Gamma$ is just the usual standard topology as a subspace of the real plane.)