TITLE: Real-analytic manifolds in real-analytic sets
QUESTION [6 upvotes]: Let $U\subset \mathbb{R}^n$ be open, and let $f:U\to\mathbb{R}$ be real-analytic. We consider the zero set $Z:=f^{-1}(\{0\})$. 
For a paper I am writing, I am looking for the best reference to the following basic fact:
If $Z$ has topological dimension equal to $d$, then $Z$ contains a real-analytic manifold of dimension $d$.

I can get this from Lojasiewicz's theorem or similar results, but that is a slightly unwieldy reference, and something probably needs to be said about how exactly one deduces it. Given that the statement is rather simple, I was wondering if someone knows of a more direct reference to this fact. 
And to add a mathematical question: This result is obviously much weaker than Lojasiewicz's theorem. Is there a proof that doesn't require developing the full structure theorem?
Many thanks for any pointers!

REPLY [3 votes]: By now, your paper is probably out, but if you need the result in the future, it can be found in:
B. Malgrange: Ideals of differential functions, Oxford University Press, 1966
In Russian translation (by A. Gabrielov, Mir 1968), this is Proposition VI.3.11:
Let $X_0$ be an analytic germ at $0 \in \mathbb{R}^n$, dim $X_0=k$. Assume that $X_0$ contains a germ $V_0$   of a variety of class $\mathcal{C}^\infty$ of dimension $k$. Then $V_0$ is a germ of an analytic variety (which is an irreducible component of the germ $X_0$).
The proof uses properties of ideals generated by germs of analytic functions. Lojasiewicz' s theorem is not used directly.