Is this 'one of' the smallest open sets that contains the original set?

Question

I have put quotations around 'one of' because the following may not be strictly the smallest, just one that satisfies my needs (and there are different ways of defining the 'size' of a set).

I am trying to prove that, for compact subsets $E \subset\mathbb{R^n}$, the Hausdorff measure, $\mathcal{H}^s(E)$, can be defined using only finite $\delta$-covers of $E$.

On my way to do this, I am trying to prove the following statement:

For all $\gamma>0$ and all $V\subset\mathbb{R^n}$, there is an open set $W$ with $V \subset W \subset \mathbb{R^n}$ and $ diam(W) \leq diam(V) +\gamma$.

Hence, my wording of this question: $W$ being the smallest such open set, since $\gamma$ is arbitrary.

Here is my work:

Let $W=V_{\frac{\gamma}{2}}=\{x \in \mathbb{R^n} \vert \exists y\in V\ $with$\ ||x-y|| < \frac{\gamma}{2} \}$. Then clearly $V \subset W$. This set can easily be shown to be open. To finish, we observe $diamW=diamV_{\frac{\gamma}{2}} = \sup_{x,y \in V_{\frac{\gamma}{2}}}||x-y|| \leq \sup_{x,y \in V}||x-y||+2\frac{\gamma}{2} = $diam$(V)+\gamma$.

I'm slightly sceptical of the last inequality. From a few rough sketches, this seems to be true, but I am not entirely convinced. Is my working correct?

In any case, the statement holds, so every cover of $E$ can be made up of open sets, and so the cover in the definition of the Hausdorff measure can be made finite. A nice little result.

Any feedback on my method for establishing $W$ would be appreciated!

Answer 1

Your argument is correct, although as you say the last inequality is a bit sketchy - you should say more about why it is true. (Think about finding points $x'$ and $y'$ close to $x$ and $y$, respectively, and which are in $V$. . .)

However, your statement

$W$ being the smallest such open set, since $\gamma$ is arbitrary

is incorrect: $W$ depends on $\gamma$ (remember that the statement is "for each $\gamma$ there is a $W$ ..."). There need be no single $W$ which works for all $\gamma$, and indeed there won't be in general: take $V$ to be the singleton containing the origin. $V$ has diameter $0$, but any open set $U$ containing $V$ has nonzero diameter; so in particular such a $U$ will "fail" for $\gamma={diam(U)\over 2}$. But for every $\gamma>0$ the open ball of radius ${\gamma\over 2}$ centered on the origin will work . . . for that specific $\gamma$.