$[a,b)$ and $(a,b)$ are not homeomorphic $[a,b)$ and $(a,b)$ are not homeomorphic
Hint given to me is: removing one point from $(a,b)$ leaves a disconnected set whereas removing a point from $[a,b)$ still gives a connected set.
I don't understand how being able to remove points and preserving connectedness while doing that is related to homeomorphism
Is there a simple way to show this?
 A: Let $f:[a,b)\to (a,b)$ be a homeomorphism. Let $x=f(a)$. Then $g:(a,b) \to (a,b) \setminus \{x\}$ is a homeomorphism. (Can you check that?). Now the range of this map is the union of the disjoint open sets $(a,x)$ and $(x,b)$ so the range is not connected. But there is a theorem which says that the image of a connected space under a  continuous function is always connected. Since the domain of $g$ is connected we have reached  a contradiction. 
A: That is, by far, the simplest way.
If the was a homeomorphism $f$ from $[a,b)$ onto $(a,b)$, then the restriction of $f$ to $(a,b)$ would then be a homeomorphism from $[a,b)$ onto $(a,b)\setminus\bigl\{f(a)\bigr\}$. But $(a,b)$ and $(a,b)\setminus\bigl\{f(a)\bigr\}$ are not homeomorphic, since $(a,b)$ s connected, whereas $(a,b)\setminus\bigl\{f(a)\bigr\}$ isn't.
A: If a property is only in terms of topology (and basic set theory and logic), then it is preserved under homeomorphisms, i.e., if $f\colon X\to Y$ is a homeomorphism, then either both $X$ and $Y$ have the property, or neither has it.
This is because $f$ is a bijection between $X$ and $Y$ and ant the same time induces an according bijection between open sets.
Now note that the property

For all $x\in X$, there exist two non-empty disjoint open subsets $U,V$ of $X$ such that $X=U\cup V\cup\{x\}$

of this kind. As $(0,1)$ has this proerty bzt $[0,1)$ does not (if we pick $x=0$), the spaces cannot be homemorphic.
