Hausdorff dimension of R x X In general, the Hausdorff dimension of a product is at least the sum of the dimensions of the two spaces. Does equality hold if one space is Euclidian?
So let $X$ be a metric space and let $\mathit{dim}_H$ denote the Hausdorff dimension. Does the following hold (and why)?
$$\mathit{dim}_H(X \times \mathbb{R}) = \mathit{dim}_H(X) + 1$$
(And if this does not hold in general, is it at least true for e.g. compact length spaces $X$?)
 A: Yes, it holds. In general it holds for any two separable metric spaces $X, Y$ with $Y$ totally bounded that
$$
\dim_H(X)+\dim_H(Y) \le \dim_H(X\times Y) \le \dim_H(X)+\dim_B(Y),
$$
where $\dim_B(Y)$ is upper box counting dimension. These inequalities can essentially be found in Falconer or Mattila's textbook (for subsets of Euclidean space, but the proofs for separable metric spaces is the same), but in short the left-most one follows from Frostman's Lemma and the right-most one directly from the definitions. Frostman's Lemma in separable metric spaces was proved by Howroyd, this is also in Mattila's book.
In particular, if $Y$ has equal Hausdorff and upper box counting dimension (which holds if $Y$ is a compact interval) then
$$
\dim_H(X\times Y)=\dim_H(X)+\dim_H(Y).
$$
Now $\mathbb{R}$ is not compact (so box counting dimension is not defined for it), but expressing $\mathbb{R}=\cup_j [-j,j]$ and using countable stability of Hausdorff dimension, it follows that the answer to your question is positive (at least when $X$ is separable - the reason I need this assumption is for Frostman's Lemma).
In fact, the inequality above holds for packing dimension $\dim_P$ in place of $\dim_B$; I stated it with $\dim_B$ because upper box counting dimension is better known, the proof of the inequality with $\dim_B$ is easier, and this is enough for answering your question. But the inequality with packing dimension is also important and is sharper (since $\dim_P\le\dim_B$).
Edit: here is the simple argument for $\dim_H(X\times [-k,k])\le \dim_H X+1$, which is the bound the OP was concerned about.
Fix $\delta>0$. Let $B(x_i,r_i)$ be a covering of $X$ with $r_i<\delta$, such that
$$
\sum_i r_i^{\dim_H(X)+\delta} < 1.
$$
(Such covering exists by definition of Hausdorff measure/dimension). For each $i$ split $[-k,k]$ into intervals $I_{ij}$ of length $2r_i$ (so there are $k/r_i$ of them). Then $B(x_i,r_i)\times I_{ij}$ has radius $\sim r_i$ and is a $O(\delta)$-covering of $X\times [-k,k]$. We obtain
$$
\sum_{ij} r_i^{\dim_H(X)+1+\delta} = 
\sum_i {k\over r_i}r_i^{\dim_H(X)+1+\delta} =
\sum_i kr_i^{\dim_H(X)+\delta}
< k.
$$
Since $k$ is fixed we obtain the estimate $\dim_H(X\times [-k,k])\le \dim_H(X)+1$.
A: Yes, taking the product with $\mathbb{R}^n$ always increases
the Hausdorff dimension by exactly $n$, and here is a reference.
In the first paragraph of [1], Kaoru Hatano notes
(in slightly different notation):

If $E \subset \mathbb{R}^m$ and $F \subset \mathbb{R}^n$
  have dimensions $\dim(E) = \alpha$ and $\dim(F) = \beta$,
  then the dimension of the product $E \times F$
  is in the range $J = [\alpha + \beta, \min(\alpha + n, m + \beta)]$.

In your case, $F = \mathbb{R}^n$,
so we get $\min(\alpha + n, m + n) = \alpha + n$,
and the range $J$ consists of just one point,
therefore $$\dim(E \times \mathbb{R}^n) = \dim(E) + n.$$
[1] Kaoru Hatano.
Notes on Hausdorff dimensions of Cartesian product sets.
Hiroshima Math. J. Volume 1, Number 1 (1971), 17-25.
http://projecteuclid.org/euclid.hmj/1206138139
