I have realized that when we are dealing with homeomorphisms in $\mathbb{R}^{n}$ we are not really considering mappings from subsets to subsets. Instead, we are constructing applications from subspaces to subspaces (where subspace means the topological space made by the subset together with the subspace topology).
When we are trying to prove that two subsets $X,Y \subset \mathbb{R}^{n}$ are not homeomorphic we usually look up for topological properties that differ between these two subsets (such as connectedness, compactness, etc.). However, when we do so, we are viewing them as subsets of $\mathbb{R}^{n}$. My first thought was that this is a mistake, because we should view them as topological spaces by themselves, not from an outside view, from an ambient space, since these properties change when we consider different topologies.
Since we do search for topological properties they have as subsets of $\mathbb{R}^{n}$ to contradict the homeomorphism possibility, there must be an explanation. I asked myself if there is a theorem that states something similar to "Let $X \subset \mathbb{R}^{n}$ be some specific topological property in the topological space $\mathbb{R}^{n}$. Then $X$ has this property in its own subspace topology". I.e, the property is invariant when we look $X$ as a subspace and as a subset of $\mathbb{R}^{n}$.
Well, this is not actually true, since a set can be neither open or close in $\mathbb{R}^{n}$, but it is in its own subspace topology. So, because we can look at them as subsets of $\mathbb{R}^{n}$ to conclude homeomorphic questions, two homeomorphic subspace need to "change equally" when "changing the point of view from $\mathbb{R}^{n}$ to subspace". How is this developed?
After all these doubts, it was a great pleasure and with an uncountable amount of joy to find the following theorem (Theorem 3.8 from "Introduction to Topological Manifolds", Springer, 2ยบ ed., by John M. Lee):
Suppose $X$ is a topological subspace and $S \subset X$ is a subspace. For any topological space $Y$, a map $f:Y \rightarrow S$ is continuous if and only if $\iota_{S} \circ f: Y \rightarrow X$ is continuous.
Bijection is not a topological property. Instead, it is a property that only uses the sets underlying the topological spaces. So, what we need to care about is the continuity of the function and its inverse. Because of the theorem above, it seems to me that we can, in fact, look at them from $\mathbb{R}^{n}$ or from themselves as topological subspaces of $\mathbb{R}^{n}$.
My final question is: Is because of the theorem above that we can do what we do (i.e, look up for contradictory topological properties of subspaces viewed from the ambient to conclude there is not homeomorphism between subspaces) ? Or could we do this since the beginning and I am missing something?
Best Answer
Well, first of all we can treat them however we want. What exactly can stop imagination, especially in maths, am I right?
Sometimes we do treat them like separate spaces, sometimes we don't. Depends on the context, on what we are trying to achieve.
But in your example I'd say that the ambient space is here only to give definition of topology for $X,Y\subseteq\mathbb{R}^n$.
Different topologies of what? Ambient space? Because $X,Y$ have to have fixed topologies, they cannot change (otherwise we would not be able to prove anything about them). So are you asking about different properties of $X,Y$ with respect to some other ambient space $Z$ they both embed to? Well, such property is not a topological property by definition. The term has a concrete meaning btw.
So for example $X$ being open in $Z$ is not a topological property. But properties of $X$ like: cardinality, separation axioms, compactness, connectedness, and so on (the list is long, see the wiki on topological property) are topological properties. If $X$ is connected, then $X$ is always connected, regardless of any space $Z$ and any embedding of $X$ into $Z$.
Being open/closed is not a topological property. In fact there is no "being open/closed" property. There is "being open/closed in [somewhere]" property. This property requires a pair "space, subspace", so of course we cannot think about them separately by definition. It just doesn't make sense.
But you are asking about "being homeomorphic". This is a very different beast and it does not depend on any ambient space.
Well, the term does not apply to topological spaces at all. It is a property of a function. Did you mean that "cardinality" is not a topological property? Of course it is. After all every homeomorphism is a bijection.
Perhaps you wanted to say that cardinality is not a complete invariant. Meaning two spaces may be equinumerous, yet they don't have to be homeomorphic. But that is not really surprising, there are lots of (all?) topological invariants which are not complete. So for example connectedness. Are any two connected subsets of $\mathbb{R}^n$ homeomorphic? Of course not. Yet, this is a topological invariant. Meaning if two spaces are homeomorphic then they have to share it.
The theorem is completely irrelevant to the topic to be honest. Yes, we could do this from the beginning. I think that the biggest mistake you've made is that you focused too much on "being open/closed" which is not even a property of a topological space (but rather a pair of spaces).