A subspace of the Cantor space

Question

Is The Cantor space $2^{\omega}$ homeomorphic to the subspace $2^{\omega} - \{(1,1,1,1,1,....)\}$ . If not, what it can be said about this subspace?

Answer 1

No, because the Cantor space is compact while $2^\omega\setminus \{(1,1,1,\ldots)\}$ is not, as the sequence $(1,0,0,\ldots),(1,1,0,\ldots,),\ldots$ converges to $(1,1,1,\ldots)$ in $2^\omega$.

Answer 2

$2^{\omega}-\{(1,1,\dots)\}$ is homeomorphic to $2^{\omega} \times \omega$ (i.e. product of Cantor set and natural numbers). See the spoiler for a homeomorphism.

Map $(\underbrace{1,1,\dots,1}_{n},0,a_1,a_2,\dots)$ to $((a_1,a_2,\dots),n)$. It is easy to see that both this map and its inverse is continuous.