Real-analytic manifolds in real-analytic sets 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!
 A: Even to say it must have some topological dimension $d$ requires some background, and it's not really that unreasonable in my opinion to go the extra mile to quote something from stratification theory like Lojasiewicz's theorem.
But you can get real-analyticity in pretty short order from just knowing the stratification is in terms of topological manifolds (I think that's what you're assuming.)  Let $D$ be a closed disk in a $d$-dimensional stratum, and let $x$ in $D$ be such that the zero of $f$ at $x$ has minimal order $a$. Then there's some multiindex $\alpha$ of order $a-1$ such that $\nabla(\partial^{\alpha}f)$ is nonzero at $x$ ($\alpha$ may be zero). Thus by the implicit function theorem the zero set of $\partial^{\alpha}f$ is an $n-1$-dimensional real-analytic manifold near $x$. On the other hand, by the minimality of $a$, $D$ is contained in this manifold (assuming the disk is small enough). So now you can look at $f$ as a function on this $n-1$ dimensional manifold, and use induction on the dimension to ultimately get that $D$ is a real-analytic manifold.
A: 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.
A: This may be a little late, but the following reference seems relevant and also
contains many other references.
@article {MR972342,
    AUTHOR = {Bierstone, Edward and Milman, Pierre D.},
     TITLE = {Semianalytic and subanalytic sets},
   JOURNAL = {Inst. Hautes \'Etudes Sci. Publ. Math.},
  FJOURNAL = {Institut des Hautes \'Etudes Scientifiques. Publications
              Math\'ematiques},
    NUMBER = {67},
      YEAR = {1988},
     PAGES = {5--42},
       URL = {67_5_0">http://www.numdam.org/item?id=PMIHES_1988_67_5_0},
}
