An Open Interval and a Half-open Interval are not Homeomorphic

Solution 1:

Without going into formalisms, if two spaces are homeomorphic then all of their topological properties are the same. Now the space $(a,b)$ has the (topological) property that if you remove any of its points, the resulting space is disconnected. But the space $[a,b)$ does not have this property -- why?

Solution 2:

Thank to all above, now I write it this way and it seems complete.

Suppose not, i.e. if there exist $f:(a,b) \to [a,b)$ a homeomorphism.

By definition, f is continuous, $f^{-1}$ is continuous, and $f$ is bijective.

$f$ is bijective then we have the inverse image of $[a,b)$ is $(a,b)$

$f^{-1}(a)$ is a single point, say c, and $a \lt c \lt b$.

Consider $f^{-1}((a,b))=(a,c) \cup (c,b)$ is disconnected.

It contradicts the fact that $f^{-1}((a,b))$ is connected, noted that $(a,b)$ connected and $f^{-1}$ is continuous.

Therefore, $(a,b)$ and $[a,b)$ are not homeomorphic.