d-dimensional topological manifold example

Question

I am having trouble understanding the concepts of manifolds, so I recently watched these two videos on the topic topology manifolds.

During watching I came up with the following example, for which I will briefly denote the required definitions I am trying to grasp first.

The definition in question given for a $d$-dimensional topological manifold is:

Definition: A topological space $(M, \mathcal{O})$ is called a d-dimensional topological manifold, if $$ \forall p \in M : \exists \mathcal{U} \in \mathcal{O} : \exists x : \mathcal{U} \to x(\mathcal{U}) \subseteq \mathbb{R}^d,$$ such that

1. $x$ is invertible: $x^{-1} : x(\mathcal{U}) \to \mathcal{U}$
2. $x$ is continuous
3. $x^{-1}$ is continuous

Note 1: In the forall condition, it is implicitly required that $p \in \mathcal{U}$.

In the video lecture, the definition for continuity is also given:

Definition: Let $(M, \mathcal{O}_M)$ and $(N, \mathcal{O}_N)$ be topological spaces. Then a map $f: M\to N$ is called continuous (w.r.t $\mathcal{O}_M$ and $\mathcal{O}_N$), if $$ \forall V \in \mathcal{O}_N : \operatorname{preim}_f(V) \in \mathcal{O}_M.$$

My own example: Let the set considered be $M = \{1, 2\}$. Of course it can be shown (and indeed it has been in the first video mentioned) that $$\mathcal{O}_M := \left\{ \emptyset, \{1\}, \{2\}, \{1, 2\}\right\}$$ is a topology on $M$. So $(M, \mathcal{O}_M)$ is a topological space.

For $\mathbb{R}^d$ we have the standard topology $\mathcal{O}_\text{standard}$ of open balls. Since I am investigating maps $x : \mathcal{U} \to x(\mathcal{U}) \subseteq \mathbb{R}^d$, I believe I should inherit a topology for $x(\mathcal{U})$ from $\mathcal{O}_\text{standard}$, which would be $$ \mathcal{O}_\text{standard}|_{x(\mathcal{U})} := \left\{ \mathcal{V} \cap x(\mathcal{U}) | \mathcal{V} \in \mathcal{O}_\text{standard} \right\}.$$ Thus the inherited topology can be directly calculated to be $$ \mathcal{O}_\text{standard}|_{x(\mathcal{U})} = \left\{ \emptyset, \{1\}, \{2\}, \{1, 2\}\right\}$$ for my example.

To determine that it is a 1-dimensional topological manifold, I investigate for each $p$ in the given set $M$:

• $p=1$: pick $\mathcal{U} \in \mathcal{O}_m$ with $p \in \mathcal{U}$ to be $\mathcal{U} = {1, 2}$ and the map $x : \mathcal{U} \to x(\mathcal{U}) = \operatorname{id}$, where obviously $x(\mathcal{U}) = {1, 2}$. Of course the identity operator is invertible, so the first condition of the definition above is met. Next, I calculate all the preimages of $\mathcal{O}_\text{standard}|_{x(\mathcal{U})}$, which happen to be

  • $\operatorname{preim}_x(\emptyset) = \emptyset \in \mathcal{O}_m$
  • $\operatorname{preim}_x(\{1\}) = \{1\} \in \mathcal{O}_m$
  • $\operatorname{preim}_x(\{2\}) = \{2\} \in \mathcal{O}_m$
  • $\operatorname{preim}_x(\{1, 2\}) = \{1, 2\} \in \mathcal{O}_m$

  to conclude $x$ is a continuous map from $(U, \mathcal{O}_M|_\mathcal{U})$ to $(x(\mathcal{U}), \mathcal{O}_\text{standard}|_{x(\mathcal{U})})$. Lastly , since the domain and the target space are identical for this example and $\operatorname{id}$ is its own inverse, nothing about the preimages will change and I conclude $x^{-1}$ is continuous, too.

• $p = 2$: With the same choice of $\mathcal{U}$ and $x$ valid, I conclude $x$ is also an invertible, continuous and $x^{-1}$ a continuous map.

Consequently, $(M, \mathcal{O}_M)$ must be a 1-dimensional topological manifold.

Is this reasoning correct? And: If I had taken $d\in \mathcal{N}$ other than 1, with $x \mapsto (x, 0, \ldots, 0)$ and $x^{-1} \mapsto x(1)$, could I have reasoned that $(M, \mathcal{O}_M)$ is any d-dimensional topological manifold?

Answer 1

So you must feel that this is not correct. Because you would get a manifold that is simultaneously $1,2,3,\ldots$ etc. dimensional. This surely doesn't sound right. In fact we know that if $M$ is an $n$-dimensional manifold then it is not a $m$-dimensional manifold for $n\neq m$.

And indeed, there is a gap in your reasoning. The very important and missing ingredient in your definition of a manifold is that $x(\mathcal{U})$ has to be an open subset of $\mathbb{R}^d$ (without a loss of generality you can even assume that $x(\mathcal{U})=\mathbb{R}^d$). A finite set of singletons is not an open subset in any $\mathbb{R}^d$ for $d\geq 1$. A singleton is open only in $\mathbb{R}^0$ (understood as a singleton $\mathbb{R}^0=\{0\}$ itself) making your $M$ a $0$-dimensional (disconnected) manifold.

Also note that you implicitely assumed that $x:\mathcal{U}\to\mathbb{R}^d$ is given by $x(1)=1$ and $x(2)=2$. Why not $x(1)=7$ and $x(2)=\pi$? What I'm saying is that you didn't define what $x$ is.

The rest seems to be correct. Good job in dealing with details. Don't take things for granted, always check yourself if possible.