is any subset of a manifold a submanifold?

by definition a submanifold is a subset of a manifold which is itself a manifold. consider $A$ a subset of an $n$-manifold $M$. a neighborhood of $x\in A$ is $\mathbb R^n$ since $x$ is an element of $M$ which is a manifold (all its elements have neighborhoods homeomorphic to $\mathbb R^n$ does this mean that every subset $A$ of $M$ is a submanifold? my guess is no and i think there is some thing wrong with my notion of neighborhood but i can't see how?

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I think there are more serious issues here than your notion of neighborhood. Think about what exactly you mean by saying "by definition a submanifold is a subset of a manifold which is itself a manifold". – t.b. Jul 9 '11 at 17:02
@Theo Buehler: Dear theo, Joriki's answer confirms that it is actually a matter of neighborhood :) – palio Jul 9 '11 at 17:56
@palio: Well, technically speaking, I was pointing out an error not in your notion of neighbourhood but in which set you applied it to -- still, if the fact that a neighbourhood is always defined with respect to some underlying set wasn't part of your notion of neighbourhood, then I guess that would qualify as an error in your notion of neighbourhood :-) – joriki Jul 9 '11 at 18:11

Whether $A$ is a manifold is determined by whether every point in $A$ has an $\mathbb R^k$-homeomorphic neighbourhood in $A$, not in $M$.

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I can think of two ways of interpreting your question, for a subset S, of a manifold M:

1) Is S, with the subspace charts of M a manifold?

Answer: no, consider, e.g., f(x)=|x| , a standardly-embedded square.

But each of these is homeomorphic to a a manifold S' ; by smoothing the corners, or, more formally, pulling back manifold charts of S.

I think most people would use this interpretation.

And:

2)Can the subset S of M be given any charts so that S is a manifold?

Edit 2: the previous example I used about a subset (not given the subspace topology)

of a manifold not being a manifold, may, at best, not be a good example. I think/hope

these cases of a subset S not given the subspace topology are better:

i) Take an uncountable subset S of $\mathbb R^n$ and give it the discrete topology.

Then S is not second countable, and we cannot modify S by a homeomorphism, into making

it a submanifold, unlike was the case of , e.g., a standardly-embedded square.

Edit1: to answer you question, and to address the key point made by Theo, S as a

subset

of M is a submanifold if the subspace charts $\{(f_i,U_i)\}$ satisfy $f(U)$=

$(x_1,..,x_k,0,0,..0)$, so that $f_i, U_i$ must be/act as the injection of $\mathbb R^n$

into $\mathbb R^k$

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Concerning 2) That's impossible: a subset of a metrizable space is metrizable (and manifolds are usually assumed to be metrizable). – t.b. Jul 9 '11 at 17:48
Theo: I actually think that a subspace S of a space X is metrizable if X is (and this is covered in point 1), but not so if S is just a subset, to which we can give different topologies, and/or where S may just not be embedded, e.g., by having self-intersections. – gary Jul 9 '11 at 18:04
Another example would be that of an uncountable subset S of $\mathbb R^n$--as a subset, not a subspace-- given the discrete topology , so that S is not 2nd-countable, and so not a manifold under any charts. I hope this does it. Or take S to be a copy/subset of the real line in $\mathbb R^2$ with the lower-limit topology. Or take the Klein bottle as a subset of $\mathbb R^n$ with n<4 . – gary Jul 9 '11 at 18:09
I honestly don't understand how your points relate to the question. We seem to agree on the main issue, however. – t.b. Jul 9 '11 at 18:12
O.K, maybe that point is not clear, or maybe just does not work, but, with the point with the uncountable discrete subspace, which is not 2nd countable, is that it cannot be given charts without losing its intrinsic discrete topology; in the first case, e.g., that of the standard embedding of the square, the issue is more the choice of embedding, and not an intrinsic property, which I think is the case with an uncountable discrete subset, in which case no embedding will do, as 2nd-countability is a topological property--I believe even preserved by continuity. – gary Jul 9 '11 at 18:21