Is every subset of $\mathbb{R}^n$ an $n+k$ topological manifold?

Question

Is every subset of $\mathbb{R^n}$ a $n+k$ topological manifold ?

It seems intuitively true because you can just decide to think of the subset as a subset of $\mathbb{R}^{n+k}$

Practically what you do is you map each point $p =(p_1,\cdots,p_n)$ to $ f(p)=\tilde{p} = (p,0,\cdots,0)$ and there's $k$ zeroes of course.

And this mapping is very analogous to the identity mapping so I guess it's bi-continuous, otherwise it would be surprising if it actually isn't.

But I just learned the definition of a topological manifold some minutes ago and I haven't yet become very comfortable with the concepts.

So if that's true how would you show it formally ? and if it's false (which I doubt) why ?

Any clarifications will be much appreciated.

Answer 1

I suppose you intend $k > 0$.

No. To be a $n+k$-dimensional topological manifold (without boundary), every point must have an open neighborhood homeomorphic to $\mathbb{R}^{n+k}$, which is not going to happen if you clamp $k$ coordinates to $0$.

Further, a single point is a subset of $\mathbb{R}^n$ and is neither an $n$-dimensional manifold nor an $n+k$-dimensional manifold, unless $n = k = 0$. The integers are a subset of $\mathbb{R}^1$ and aren't even a $0$-dimensional manifold (because they are not Hausdorff in the induced topology).

Another way to come at this is that a neighborhood of a point in an $n$-dimensional manifold has positive measure using the $n-$fold product of Lebesgue measures.

(If we want to talk about manifolds with boundary and points on the boundary, each has a neighborhood homemorphic to the half space $\mathbb{R}^n$ with last coordinate $\geq 0$, so has neighbors with $n$-dimensional neighborhoods and has a neighborhood with positive $n$-dimensional measure.)

Answer 2

the definition you later work with will probably look like this: A $n$-dimensional manifold is a set $M$ with an open cover $\{U_i\}_I$ for which there exists homeomorphisms $\phi_i:U_i \rightarrow \mathbb{R}^n$ for each $i \in I$. The intuition you keep in mind is that you think of $M$ as an object that looks locally like $\mathbb{R}^n$.

Answer 3

What you have from that is that every subset of $\mathbb{R}^n$ has a homeomorphic copy in $\mathbb{R}^{n+k}$ for $k\geq0.$