Are these subsets of $\mathbb{R}^2$ homeomorphic? The spaces are $\mathbb{R} \times [0,1]$ and $\mathbb{R} \times (-\infty,0]$.
So, if [0,1] was either [0,1) or (0,1] showing homeomorphism is trivial. However, the inclusion of both the points leads me to believe that there will not even exist a (surjective) continuous map between the two spaces.
 A: No. The only proof I can come up with uses some algebraic topology:
Both $\mathbb{R}\times[0,1]$ and $\mathbb{R}\times(-\infty,0]$ are topological manifolds with boundaries. The boundary of $\mathbb{R}\times[0,1]$ consists of two lines, and is disconnected. The boundary of $\mathbb{R}\times(-\infty,0]$ is a line, and connected.
However, any homeomorphism between topological manifolds with boundaries restricts to a homeomorphism between their boundaries; see


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*Do homeomorphic manifolds with boundary have homeomorphic interiors?,


(the boundary of a manifold is the complement of its interior) so those spaces are not homeomorphic.
A: I think Luiz's answer is right on the money, but here is an alternative anyway. If these spaces were homeomorphic, then so would be their 1-point compactifications. However, the 1-point compactification of $\mathbb{R} \times [0,1]$ is homotopy equivalent to the circle $S^1$, whereas the 1-point compactification of $\mathbb{R} \times (-\infty,0]$ is contractible (it is homeomorphic to the closed 2-disk). So, as long as you have enough tools to know that the circle is not contractible, you can apply these observations to conclude that the spaces are not homeomorphic.
A: Here's an approach that doesn't use any algebraic topology; you only need Heine-Borel. If you remove a compact subset from $\mathbb{R} \times (-\infty,0]$ the remainder has only one (connected) component whose closure is not compact. If you do the same to $\mathbb{R} \times [0,1]$ the remainder can have two such components.
