Let $\mathcal{H}^n$ denote the Hausdorff measure, $\dim\_H X$ the Hausdorff dimension, and $\dim X$ the topological dimension of $X$. 


A well known result of
Szpilrajn (He changed his name to Marczewski while hiding from Nazi persecution) proved in [S] asserts that
if $\mathcal{H}^{n+1}(X)=0$, then the topological dimension of $X$ is at most $n$. 


Szpilrajn's proof is reproduced in [Theorem 7.3, HW] and [[Theorem 8.15](https://books.google.com/books?id=5oXlBwAAQBAJ&pg=PA63), H]. 


Szpilrajn however, mentions in [S] that his argument is based on Nöbeling's proof of a weaker result that topological dimension is bounded from above by the Hausdorff dimension of a metric space. However, he did not provide any reference to Nöbeling's work. 


There is also no reference to Nöbeling's work in the book by Hurewicz and Wallman. 



> 
> **Question.** Does anybody know the reference to the original work of Nöbeling?
> 
> 
> 


**[H] J. Heinonen,**
*Lectures on analysis on metric spaces.* Universitext. Springer-Verlag, New York, 2001.


**[HW] W. Hurewicz, H. Wallman,** *Dimension Theory.* Princeton Mathematical Series, v. 4. Princeton University Press, Princeton, N. J., 1941.


**[S] E. Szpilrajn,** [La dimension et la mesure](http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-fmv28i1p11bwm), *Fund. Math.* 28 (1937), 81--89.



Unknown work of Nöbeling on topological/Hausdorff dimension
So, the sought for paper is:


Nöbeling, G., Hausdorffsche und mengentheoretische Dimension, Ergebnisse math. Kolloquium Wien 3, 24-25 (1931).


And here is a ``translation" (to English and to modern math exposition standards, if such a thing exists.) What is called the set-theoretical dimension is defined inductively: dim of the empty set is $-1$, and we set $\dim M = k$ if $k$ is the least integer with the property that every point of $M$ has arbitrarily small (open) neighborhoods whose dim is $k-1$.


Hausdorff and set-theoretical dimension. By George Nöbeling.
------------------------------------------------------------


Let $M$ be a subset of the Euclidean $\mathbb{R}^n$. One covers $M$ by finitely or countably (infinitely) many balls $K\_j$, with diameters $d\_j < \rho$, for $p \leq n$ (any non-negative number) form the sum $ \sum d\_j^p $. Let $ L\_p(\rho, M)$ be the infimum of such sums for all such coverings. Put 
$$
L\_p (M) = \lim\_{\rho \to 0} L\_p (\rho, M) .
$$
Obviously there is exactly one number $p= p (M)$ such that for every $q > p$, $L\_q (M) = 0$ and for every number $q <p$, $L\_q (M) = \infty$. We call this clearly defined number $p$ the Hausdorff dimension of the set $M$ and claim that


**THEOREM**: For any set $M$, the Hausdorff dimension is at least equal to the set-theoretical dimension.


**PROOF.** The proof is done by induction on the set-theoretical dimension $\dim M$. For every $k \in \{0,1,2,\cdots\} $ we prove that if $M$ is a set with $\dim M \geq k$ then $p(M) \geq k$.


For $k=0$ this is obvious thanks to $ p (M) \geq 0.$ Suppose the claim is true for $k$. We must thus prove that if $M$ is any set with $\dim M \geq (k + 1)$, then $p(M) \geq k+1$.


Since $M$ is at least $(k+1)$-dimensional, there exit a point $P$ of $M$ and a number $r\_0$, such that for every $(n-1)$-dimensional sphere $S\_r$, with the radius $r \leq r\_0$ the intersection $M \cap S\_r$ is at least $k$-dimensional. [*Otherwise, every point of $M$ would have arbitrarily small open neighborhoods whose boundaries had dim $k-1$ or less, and so by definition $M$ would have dimension less than or equal to $k$.*], and therefore according to the induction hypothesis that
$$
\forall r \leq r\_0, \; p (M \cap S\_r) \geq k \, .
$$ 


For each $i \in \mathbb{N}$, let $\{K\_{ij}\}\_j$ be a covering of the set $M$ by spheres of diameter $d\_{ij} < \frac{1}{i}$. For a number $q < k +1$ and an $0 < x \leq r\_0$ we set 
 $$
f\_{ij} (x) = 
\begin{cases}
d\_{ij}^{q-1} & \text{if $S\_x \cap K\_{ij} \neq \emptyset $ ,}\\
0 & \text{Otherwise.}\\
\end{cases}
$$


We also set 
$$
s\_i (x) = \sum f\_{ij} (x).
$$
Obviously,
$$
s\_i(x) \geq L\_{q-1}(1/i,S\_x \cap M) \, .
$$
Since $ p (M \cap S\_r) \geq k > q-1 $, it follows from the induction hypothesis that
$$
\forall x \in (0,r\_0], \; \lim\_{i \to \infty} s\_i(x) = \infty \, .
$$


Thus,
$$
 \sum\_j \int\_0^{r\_0} f\_{ij}(x) \, dx = \int\_0^{r\_0} s\_i(x) \, dx \xrightarrow{i \to \infty} \infty \, .
$$
Now observe that $f\_{ij} (x) = d\_{ij}^{q-1}$ for $x$ from an interval whose length is at most $d\_{ij}$ -- the diameter of $K\_{ij}$ -- and otherwise $f\_{ij} (x) = 0.$ Therefore, 
$$
\sum\_j d\_{ij}^{q} = \sum\_j \int\_0^{d\_{ij}} d\_{ij}^{q-1} \, dx \geq \sum\_j \int\_0^{r\_0} f\_{ij}(x) \, dx \, .
$$
and hence,
$$
\sum\_j d\_{ij}^{q} \xrightarrow{i \to \infty} \infty.
$$
Since this is true for any coverings $K\_{ij}$, we conclude
$$
L\_q(M) = \infty \implies p(M) \geq q \, .
$$


Since $q <k + 1$ was arbitrary we have shown that $p (M) \geq k + 1$. This concludes the induction and proves our theorem. $\Box$


