Is every open subset of a manifold also a manifold?

If $M$ is a manifold, is it true that every open subset of $M$ is a manifold under the subspace topology?

My definition for manifold is, for every point $x \in M$ there is a open set $U \subset M$ such that $x \in U$ and $U$ is homeomorphic to the unit ball in $\mathbb{R}^n$ for some $n$, and $M$ is Hausdorff.

I was figuring that this would be the case since an open subset shouldn't hurt the points inside locally too much. I am not sure how to go about proving this though.

• possible duplicate of An open subset of a manifold is a manifold – Dan Rust Jun 13 '13 at 13:52
• @DanielRust I don't think this is a duplicate since the definitions are different. – Tom Oldfield Jun 13 '13 at 13:59
• In the linked question can he say $h(A \cap U_x)$ is open since $A \cap U_x$ is open, and $h$ is a homeomorphism? With my definition I believe I can still say its open, then I can take an open ball around $h(x)$ which would be homeomorphic to the unit ball. I think. Then the pull back of that open ball would be the open set around $x$ I need. – user82241 Jun 13 '13 at 14:14
• @Daniel, Aha, thank you! I'll delete that comment then. – Tom Oldfield Jun 13 '13 at 14:14

Yes. Let $U$ be an open subset of a manifold $M$.
Suppose $x \in U \subset M$. We know there is some open set $O \subset M$ containing $x$ and a homeomorphism $\phi: O \to B_1(\mathbb{R}^n)$. Here $O$ has the subspace topology.
Since $U$ is open, and $\phi$ is a homeomorphism, the image of the open set $O \cap U$ under $\phi$ is open in $B_1(\mathbb{R}^n)$. Now, $\phi(O \cap U)$ also contains $\phi(x) = 0$, so it must contain an entire ball $B_\epsilon(\mathbb{R}^n)$ for some $\epsilon > 0$.
Let $\tilde{O} = \phi^{-1}(B_\epsilon(\mathbb{R}^n)) \subset U$, and note that $\tilde{O}$ is an open neighborhood of $x$ both in $M$ and in $U$ with the subspace topology. Finally, $\psi: z \in \tilde{O} \mapsto \frac{1}{\epsilon}\phi(z)$ is a homeomorphism between $\tilde{O}$ and $B_1(\mathbb{R}^n)$.