Let $M\subseteq \mathbb{R}^k$: Manifold topology vs. trace topology I'm confused about the topology of submanifolds of $\mathbb{R}^n$: Let $M$ be such a $k$-manifold (say, the circle $S^1$, of dimension $1$, embedded in say $\mathbb{R}^7$); the topology of such a manifold is defined by the initial topology of all the charts $\varphi_i:  U\rightarrow \mathbb{R}^k$.
To be more precise, what I'm confused about is, what the relationship between the topology on  $M$ as defined above is (i.e. the "manifold topology") in contrast to the trace topology $M$ as a subset of $\mathbb{R}^k$, inheriting the standard topology on $\mathbb{R}^k$.
In particular I'm interested in understanding what it means for $M$ to be compact. If the two topologies mentioned above were the same, than that would mean that compact $M$, as a subset of $\mathbb{R}^k$, fulfills the Heine-Borel characterization of compactness, i.e. a compact manifold is bounded and closed - which is what I ultimately would like to know.
 A: (Edit: I am making this answer longer since I now have a keyboard.)
There is some definition confusion in your question.
To say that $M$ is a submanifold of $ \mathbb {R}^n $ is to say that these topologies agree.
To be more precise: to say that $M$ is a manifold means that $M$ is a topological space equipped with charts. The charts are required to be continuous for the topology that is already on $M$.
To say that $M$ is a submanifold of $\mathbb{R}^n$ is to say that $M$ is a manifold with the subspace topology coming from $\mathbb{R}^n$, and so, by definition, for a submanifold, these topologies agree. If the topology coming from the charts "didn't agree with" the subspace topology, then the charts wouldn't define a manifold structure on the topological space $M$ but rather would define a manifold structure on some other topological space which happens to have the same underlying set as $M$.
For example, take the open unit interval in $\mathbb{R}^2$. With its subspace topology, it's a submanifold. 
But we could pick a (totally non-continuous) bijection between the open unit interval and the square $(0, 1) \times (0, 1)$. Then we could use that topology to equip the interval itself with a manifold structure coming from the standard manifold structure on the square. That's a subset of $\mathbb{R}^2$, which we are considering as a manifold, but it is certainly not a submanifold of $\mathbb{R}^2$, because it's not even a subspace of $\mathbb{R}^2$; just some subset that you're endowing with an irrelevant topology.
