if there exist a topological space such that we can not find a homeomorphism mapping to find its open cover?

Question

The definition of the manifold in the textbook is For a topological space $M=(X,T)$, and for a set of the subset $\{O_a\}$ that for all $a$, $O_a\in T$ such that

1. $\{O_a\}$ is an open cover of $M$;
2. For all $O_a$, exists homeomorphism map $\Phi_a: O_a\rightarrow V_a$; Add: $V_a$ is an open subset measured by the usual topology in $\mathbb{R}^n$.
3. if for all $O_a\cap O_b\ne \emptyset$, then $\Phi_b \circ\Phi_a^{-1}$ is $C^{\infty}$;

Then $\{O_a\}$ is the manifold of $M$.

And here is an example: $$M=\mathbb{R}^2$$ where the topology of $M$ is $\{\emptyset, \mathbb{R}^2\}$ (usual topology). Then we can say the open cover of $M$ is given by $$\{\emptyset, \mathbb{R}^2\}$$ (This satisfies the Condition 1)

And there exists a homeomorphism map from $O_a=\mathbb{R}^2$ ***to a specific open subset in *** $\mathbb{R}^2$, condition 2 holds;

And we can let $O_b=\emptyset$ without loss of generality. then $O_a\cap O_b\ne \emptyset$, which means this identity map is smooth, and condition 3 holds. Thus one of the $M$'s manifolds is $\{\mathbb{R}^2\}$...

I am just wondering if there are any examples of $M$ (topological space) such that we can not find a homeomorphism mapping to satisfy condition 2. I would appreciate it for someone set a counterexample for me.

=========Edit========

This is similar to my question:"https://math.stackexchange.com/q/4544650/1095737", where the answer said: "So you prove that a topological space is not a manifold by proving that one of these conditions fails for a given topology".

Yes, I am just wondering if there are any examples that, when proofing an open set that is not manifold, could we show that there is not exist homeomorphism mapping (condition 2)? So it is close to my question. But I would ask one more about condition 2: can we use some methods to show homeomorphism mapping is not exists?

I agree with Gribouillis's comments, and thanks for your help. So for the set of all rational numbers $\mathbb{Q}$, let $\{O_q\}$ be the open cover of $\mathbb{Q}$, and we use condition 2, to find if there is a mapping such that $O_q\rightarrow V_q$ where $V_q$ which is an open subset measured by the usual topology in $\mathbb{R}^n$...

But the mapping from $\mathbb{Q}$ to $\mathbb{R}^n$ I think does exist. In fact, a mapping can be constructed from the set of all rational numbers to the set of all real numbers by mapping each rational number to its corresponding position on the real number line. This mapping is typically referred to as the inclusion map or indicator function on the real numbers, which maps each rational number to its corresponding point on the real number line and maps each non-rational real number to a distinct point on the real number line.

Therefore, even though rational numbers can be represented as fractions, they are still a part of the set of real numbers and can be mapped one-to-one with the real numbers through a mapping.

So the example of $\mathbb{Q}$ I personally think cannot be set as an example to use condition 2 to judge whether it is a manifold or not. (If I made some mistake, please do not hesitate to point it out).

======Update======

Thanks to Mr. Elchanan Solomon. The example I set above is not the 1-1 mapping. In condition 2, this homeomorphism map should be one-one onto mapping. So for the example of the set of all rational numbers, I think I cannot give or find such a mapping that satisfies condition 2. Really appreciate!

Answer 1

抱歉如果我没有清楚地懂你问题的意思。

Not sure if this answers your question, but let me try and say something that might be helpful. A manifold is a topological space that looks, locally, like Euclidean space. Therefore, a manifold must have the same local topological properties as Euclidean space.

One such property is being locally connected. The rationals $\mathbb{Q}$ are not locally connected as a topological space (in fact, they are totally disconnected), so they cannot be a manifold. You might argue that the rational numbers can be mapped to the real numbers in a one-to-one way. This is both false and misleading. First of all, there is no such one-to-one map, this is a classic result in Set theory. Secondly, when we talk about $\mathbb{Q}$ as a topological space we are talking about the topology it inherits as a subset of $\mathbb{R}$, not any arbitrary topology you decide to equip it with.

Another property is being locally compact. To make a non locally compact space, take infinitely many copies of the unit interval $[0,1]$ and glue them together at $0$. This cannot be given a manifold structure.