Are X and Y homeomorphic? Let X=(0,1) in R and Y=(3,23) in R be equipped with the subspace topology of (R, Teuclidean). Are X and Y homeomorphic?
I am not too sure where to go with this, I was trying to follow an example in Topology Without Tears but I don't really understand why it is doing what it is doing.
So, so far I have;
Let (3,23) in R with 3<23, and consider the function f: (0,1) -> (3,23) given by 
f(x)=3(1-x)+23x.
But I don't know what that is about or how to progress further.
Thanks
 A: You can always match two intervals one on top of the other linearly.
If $(a,b)$ and $(c,d)$ are two intervals then
$$ f(t) = \frac{c-d}{a-b}t + c - a\frac{c-d}{a-b}$$
gives such a transformation.
Note that $f$ is automatically continuous (for example because it's differentiable).
Let's check what happens in your example: $(a,b) = (0,1)$ and $(c,d) = (3,23)$
so the corresponding $f$ is
$$ f(t) = -20t + 3 = 3(1-t) + 23t.$$
What's left to check that it is a homeomorphism is that it is bijective with continuous inverse.
Let's apply the formula above but this time with $(a,b) = (3,23)$, $(c,d)=(0,1)$.
We have
$$ g(s) = \frac{-1}{-20}s-3\frac{-1}{-20} =\frac{1}{20}\left( s - 3 \right).$$
If we can show that $g$ is the inverse of $f$ then we are done.
To show that, one needs to prove both $g(f(t))=t$ and $f(g(s))$.
This can be done simply by plugging in.
The same argument shows that any two intervals are homemorphic provided they are of the same kind (open, closed, semi-open). To be pedantic, to show that $[a,b)$ is homeomorphic to $(c,d]$ one needs to tweak our transformations so that they have negative linear part.
A: Two spaces are homeomorphic (by definition) if there is a continuous, bijective function $f$ between them, such that the inverse is also continuous and bijective. 
So this problem has already proposed such a map, and it remains to show that it is continuous and bijective, and then, after writing down the inverse, showing that this is continuous and bijective also.
So you have $6$ things to prove, and $1$ thing to calculate:
(1) $f$ is continuous
(2) if $f(x_1) = f(x_2)$, then $x_1 = x_2$
(3) for every $y \in (3,23)$, there is an $x \in (0,1)$ so that $f(x) = y$
Calculate the inverse (one way is to write $y = 3(1-x) + 23x$, then solve for $x$), then prove (1)-(3) for it (with the statements modified as appropriate)
