Is there exist a homeomorphism between either pair of $(0,1),(0,1],[0,1]$ As the topic is there exist a homeomorphism between either pair of $(0, 1),(0,1],[0,1]$
 A: No. The idea of the proof is that when you remove a point from $(0,1)$, you end up with a disconnected space. So first we shall show that $(0,1)\not\cong [0,1)$. Assume that $f:[0,1)\to (0,1)$ is a homeomorphism. Then, remove the point $\{0\}$ from the domain. That is $A=[0,1)-\{0\}$ meaning that $$f\vert_{A}:(0,1)\to (0,1)-\{f(0)\}$$ is also a homeomorphism since $(0,1), [0,1]$ and $[0,1)$ are all in the subspace topology of the usual topology on $\mathbb{R}$. However this is a continuous bijection from a connected to a disconnected space, which is impossible! Now I will show that $(0,1)\not\cong [0,1]$. We use a similar idea; that is we remove $\{0\}$ from $[0,1]$ meaning that if we have a homeomorphism $\varphi:[0,1]\to (0,1)$, then the restriction of this homeomorphism to $(0,1]$ $$\varphi\vert_{(0,1]}:(0,1]\to (0,1)-\{f(0)\}$$ is a continuous bijection from a connected to a disconnected space, a contradiction again! Finally we move onto the most difficult part of the proof: showing that $[0,1]\not\cong [0,1)$. The idea is the same, but it's slightly harder to execute. The idea is that when we remove $0$ from both spaces, we get $(0,1)$ and $(0,1]$ and then when we remove $1$ from both spaces, we get a disconnected space $(0,1)-\{1\}$ and a connected space $(0,1)$ and we use the same idea. If we remove any other points from $(0,1]$, we get a disconnected space. So overall we have it that if there were a homeomorphism $f:[0,1]\to [0,1)$, then there would be a homeomorphism $$f\vert_{[0,1]-\{0,1\}}(0,1)\to [0,1)-\{f(0),f(1)\}$$ which is a continuous bijection from a connected to a disconnected space which is impossible
A: No two of the three spaces are homeomorphic. One way to see this is to note that $(0,1)$ has no non-cut points, $(0,1]$ has one non-cut point, and $[0,1]$ has two. (A non-cut point is one whose removal does not disconnect the space.) Another way to see that $[0,1]$ is not homeomorphic to either of the others is to note that $[0,1]$ is compact, and they are not. $(0,1)$ and $(0,1]$ can also be distinguished by the fact that the one-point compactification of $(0,1)$ is homeomorphic to the circle $S^1$, while that of $(0,1]$ is homeomorphic to $[0,1]$.
