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A Manifold 'Properly' Contained in Another

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Question: Let $M$ be a $k$-manifold-without-boundary in $\mathbb R^n$ and $N$ be another manifold-without-boundary in $\mathbb R^n$ such that $M\subseteq N$.


Assume that there exists a point $\mathbf p\in M$ such that each neighborhood $U$ of $\mathbf p$ has a point $\mathbf q\in U$ such that $\mathbf q\in N\setminus M$.


Then can $N$ possibly be a $k$-manifold?
___


Intuitively it seems obvious that the dimension of $N$ should be greater than $k$ but I haven't been able to make any progress to make it into a proof.


Can somebody help?


Thanks.
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Question: Let $M$ be a $k$-manifold-without-boundary in $\mathbb R^n$ and $N$ be another manifold-without-boundary in $\mathbb R^n$ such that $M\subseteq N$.


Assume that there exists a point $\mathbf p\in M$ such that each neighborhood $U$ of $\mathbf p$ has a point $\mathbf q\in U$ such that $\mathbf q\in N\setminus M$.


Then can $N$ possibly be a $k$-manifold?
One of my professors came up with a nice short argument. I will lengthen it by a lot to add more detail.

We will work more generally with topological manifolds.

Definition: Let $M\subseteq \mathbb{R}^n$. We say that $M$ is a topological $k$-manifold if for every $p\in M$ there exists an open set $U\subset \mathbb{R}^n$, an open set $V\subset \mathbb{R}^k$ and a homeomorphism $\varphi: (M\cap U)\to V$. "Homeomorphism" is just that $\varphi$ is bijective, continuous, and $\varphi^{-1}$ is continuous.

Notice this is essentially the same exactly definition as smooth manifold, except instead of $\varphi$ being a hemeomorphism it is a diffeomorphism. We will prove the following theorem. I will assume you know some basic topological definitions.

Theorem: Let $M\subseteq N \subseteq \mathbb{R}^n$ be topological $k$-manifolds, in this case $M$ is open subset of $N$.

Recall, that $N$ being a subset of $\mathbb{R}^n$ inherits the topology of $\mathbb{R}^n$ in the following way. The open subsets of $N$ are defined by $W\cap N$ where $W$ is an open subset of $\mathbb{R}^n$. Thus, when we claim that $M$ is an open subset of $N$ we are claiming that $M = W\cap N$ for some open set $W\subseteq \mathbb{R}^n$.

Corollary: If $M\subseteq N\subseteq \mathbb{R}^n$, topological manifolds, and there is a $p\in M$ such that for any open set $U\subseteq \mathbb{R}^n$ containing $p$ there is a point $q\in U$ with $q\in N\setminus M$ then $\dim M < \dim N$.

Proof: If $\dim M = \dim N$ by theorem $M$ would be an open subset of $N$. In particular, there be an open set $U\subseteq \mathbb{R}^n$ such that $M = (U\cap N)$ is open. Now it follows that for every $p\in M$ we have $p\in U$ and since $N\cap U = M$ we have that every $q\in U\cap N$ must belong to $M$ as well which contradicts our hypothesis.

Let us prove the theorem. The theorem rests on the following topological result.

Brouwer's Invariance of Domain: Let $U\subseteq \mathbb{R}^n$ be an open set and $f:U\to \mathbb{R}^n$ be a continuous injective map. Then $f$ is an open map. In other words, for any open set $V$, a subset of $U$, we have $f(V)$ is open, so in particular, $f(U)$ is open.

Proof: Let $p\in M$, since $M$ and $N$ are both $k$-manifolds we can find an open sets $A,B\subseteq \mathbb{R}^n$, each containing $p$, open sets $X,Y\subseteq \mathbb{R}^k$, and homeomorphisms $\varphi: (A\cap M)\to X, ~ \psi: (B\cap N)\to Y$. Let $U = A\cap B$, so $U$ is an open set. Let $V = \varphi(U)$ and $W = \psi(U)$, these are open subsets of $\mathbb{R}^k$ since $\varphi,\psi$ are homeomorphism (so it maps open sets to open sets). Thus, we have homeomorphisms $\varphi|_U: (U\cap M)\to \varphi(U)$ and $\psi|_V: (V\cap M)\to \psi(V)$. [There is a slight abuse on notation when these maps are restricted but I hope you understand that was only done to simplify the notation.]

Basically, all we are trying to say, is that, for $p\in M$ we can find an open $U\subset \mathbb{R}^n$, open $V,W\subset \mathbb{R}^k$, with $p\in U$, and homeomorphisms $\varphi: (U\cap M)\to V$ and $\psi: (U\cap N)\to W$.

Let $i: (U\cap M)\to (U\cap N)$ be the inclusion map $i(x) = x$ which is well-defined and continuous as $M\subseteq N$. Define $F=\psi \circ i\circ \varphi^{-1}:V\to W$. This map is injective as each function comprosing it was. This map is continuous as each in it is continuous. Furthermore, $V,W\subset \mathbb{R}^k$ so that $F:V\to \mathbb{R}^k$ and it satisfies the conditions of IOD theorem, it follows that $F$ is an open map.

We claim that $i: (U\cap M)\to (U\cap N)$ is an open map as well. Pick any $\mathcal{O}$ open in $(U\cap M)$. Then $i(\mathcal{O}) = \psi^{-1}( F(\varphi(\mathcal{O})))$ which is open, as each map $\psi^{-1},F,\varphi$ in the composition is open. Therefore, $U\cap M$ is an open subset of $U\cap N$.

I leave the last step to you to figure out why this implies that $M$ must be open in $N$.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
One of my professors came up with a nice short argument. I will lengthen it by a lot to add more detail.

We will work more generally with topological manifolds.

Definition: Let $M\subseteq \mathbb{R}^n$. We say that $M$ is a topological $k$-manifold if for every $p\in M$ there exists an open set $U\subset \mathbb{R}^n$, an open set $V\subset \mathbb{R}^k$ and a homeomorphism $\varphi: (M\cap U)\to V$. "Homeomorphism" is just that $\varphi$ is bijective, continuous, and $\varphi^{-1}$ is continuous.

Notice this is essentially the same exactly definition as smooth manifold, except instead of $\varphi$ being a hemeomorphism it is a diffeomorphism. We will prove the following theorem. I will assume you know some basic topological definitions.

Theorem: Let $M\subseteq N \subseteq \mathbb{R}^n$ be topological $k$-manifolds, in this case $M$ is open subset of $N$.

Recall, that $N$ being a subset of $\mathbb{R}^n$ inherits the topology of $\mathbb{R}^n$ in the following way. The open subsets of $N$ are defined by $W\cap N$ where $W$ is an open subset of $\mathbb{R}^n$. Thus, when we claim that $M$ is an open subset of $N$ we are claiming that $M = W\cap N$ for some open set $W\subseteq \mathbb{R}^n$.

Corollary: If $M\subseteq N\subseteq \mathbb{R}^n$, topological manifolds, and there is a $p\in M$ such that for any open set $U\subseteq \mathbb{R}^n$ containing $p$ there is a point $q\in U$ with $q\in N\setminus M$ then $\dim M < \dim N$.

Proof: If $\dim M = \dim N$ by theorem $M$ would be an open subset of $N$. In particular, there be an open set $U\subseteq \mathbb{R}^n$ such that $M = (U\cap N)$ is open. Now it follows that for every $p\in M$ we have $p\in U$ and since $N\cap U = M$ we have that every $q\in U\cap N$ must belong to $M$ as well which contradicts our hypothesis.

Let us prove the theorem. The theorem rests on the following topological result.

Brouwer's Invariance of Domain: Let $U\subseteq \mathbb{R}^n$ be an open set and $f:U\to \mathbb{R}^n$ be a continuous injective map. Then $f$ is an open map. In other words, for any open set $V$, a subset of $U$, we have $f(V)$ is open, so in particular, $f(U)$ is open.

Proof: Let $p\in M$, since $M$ and $N$ are both $k$-manifolds we can find an open sets $A,B\subseteq \mathbb{R}^n$, each containing $p$, open sets $X,Y\subseteq \mathbb{R}^k$, and homeomorphisms $\varphi: (A\cap M)\to X, ~ \psi: (B\cap N)\to Y$. Let $U = A\cap B$, so $U$ is an open set. Let $V = \varphi(U)$ and $W = \psi(U)$, these are open subsets of $\mathbb{R}^k$ since $\varphi,\psi$ are homeomorphism (so it maps open sets to open sets). Thus, we have homeomorphisms $\varphi|_U: (U\cap M)\to \varphi(U)$ and $\psi|_V: (V\cap M)\to \psi(V)$. [There is a slight abuse on notation when these maps are restricted but I hope you understand that was only done to simplify the notation.]

Basically, all we are trying to say, is that, for $p\in M$ we can find an open $U\subset \mathbb{R}^n$, open $V,W\subset \mathbb{R}^k$, with $p\in U$, and homeomorphisms $\varphi: (U\cap M)\to V$ and $\psi: (U\cap N)\to W$.

Let $i: (U\cap M)\to (U\cap N)$ be the inclusion map $i(x) = x$ which is well-defined and continuous as $M\subseteq N$. Define $F=\psi \circ i\circ \varphi^{-1}:V\to W$. This map is injective as each function comprosing it was. This map is continuous as each in it is continuous. Furthermore, $V,W\subset \mathbb{R}^k$ so that $F:V\to \mathbb{R}^k$ and it satisfies the conditions of IOD theorem, it follows that $F$ is an open map.

We claim that $i: (U\cap M)\to (U\cap N)$ is an open map as well. Pick any $\mathcal{O}$ open in $(U\cap M)$. Then $i(\mathcal{O}) = \psi^{-1}( F(\varphi(\mathcal{O})))$ which is open, as each map $\psi^{-1},F,\varphi$ in the composition is open. Therefore, $U\cap M$ is an open subset of $U\cap N$.

I leave the last step to you to figure out why this implies that $M$ must be open in $N$.
Thank you so so much for taking your time out to post this long answer. Really.

You have proved something stronger than what I required!

I think I should be able to simplify the argument by using the fact that $M$ and $N$ are not just topological manifolds but also differentiable manifolds. I may not need to use 'Brower's Invariance of Domain'. I can use exactly what is used in advanced calculus books to prove the Inverse Function Theorem.
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Thank you so so much for taking your time out to post this long answer. Really.

You have proved something stronger than what I required!

I think I should be able to simplify the argument by using the fact that $M$ and $N$ are not just topological manifolds but also differentiable manifolds. I may not need to use 'Brower's Invariance of Domain'. I can use exactly what is used in advanced calculus books to prove the Inverse Function Theorem.
You do not need to use IOD theorem. You just need the smooth version:

Theorem: Let $U\subseteq \mathbb{R}^n$ be open and $f:U\to \mathbb{R}^n$ is smooth and injective then $f(U)$ is open.

The proof above goes through exactly for smooth manifolds by replacing the words "homeomorphic" with "diffeomorphic".