Proof for Steinhaus theorem I try a very long time to understand the proof of this theorem from here: https://en.wikipedia.org/wiki/Steinhaus_theorem
I'm pretty sure it's wrong - the part where they say that $K+V\subset U$ just can't be true.
For example if I take $A=[0,1]$ and $U=(-0.1,1.1),\, K=[0.1,1]$ and I will take the cover $V_1=(-0.5,0.1)$ with $k_1=0.4$ and $V_2=(-0.4,0.3)$ with $k_2=0.8$ so $-0.2\in V$ and $0.1\in K$ but $-0.2+0.1\notin U$
Am I wrong or there is something I'm missing?
 A: I followed the reference on Wikipedia to the note by Karl Stromberg, who in turn cites the book Abstract Harmonic Analysis by Hewitt and Ross for the fact that we may find the neighborhood $V$ of $0$ such that $K+V\subset U$. The proof in Hewitt and Ross is very elementary and quite similar to the incorrect proof on Wikipedia, so I edited the wikipedia page to match the idea of that proof. Here's the relevant part:
Since $K\subset U$, for each $k\in K$, there is a neighborhood $W_k$ of $0$ such that $k+W_k\subset U$, and, further, there is a neighborhood $V_k$ of $0$ such that $2V_k \subset W_k$. For example, if $W_k$ contains $(-\epsilon,\epsilon)$, we can take $V_k = (-\epsilon/2,\epsilon/2)$. The family $\{k+V_k \mid k\in K\}$ is an open cover of $K$. $K$ is compact, hence one can choose a finite subcover $\{k_1+V_{k_1},\dots,k_n+V_{k_n}\}$. Let $V=V_{k_1}\cap\dots\cap V_{k_n}$. Then,
\begin{align}
K+V&\subset ((k_1+V_{k_1}) \cup\dots\cup (k_n+V_{k_n})) + V \\
&\subset ((k_1+2V_{k_1}) \cup \dots \cup (k_n+2V_{k_n}))\\
& \subset ((k_1+W_{k_1}) \cup \dots \cup (k_n+W_{k_n}))\\
& \subset U
\end{align}
A: Assuming that that $k_1,k_2$ and $V_1$,$V_2$ come from the subcover of the cover formed by all the $k+V$, for $k\in K$, you raise a valid point, the current Wikipedia proof is incorrect, as your example shows. 
To make this post self-contained (as Wikipedia's proof may end up been edited, partly as a result of this post), here is an excerpt of what they currently have. 
$K\subset U \subset \mathbb R$, where $K$ is compact, $U$ is open.
"Since $K\subset U$, for each $k\in K$, there is a neighborhood $V_k$ of $0$ such that $k+V_k\subset U$. The family $\{k+V_k : k\in K\}$ is an open cover of $K$. $K$ is compact, hence one can choose a finite subcover $\{k_1+V_{k_1},\dots,k_n+V_{k_n}\}$. Let $V:=V_{k_1}\cap\dots\cap V_{k_n}$. Then $K + V ⊂ U$."
As your example shows, the above reasoning is incorrect. 
The imprecise part of the Wikipedia proof could be fixed by taking into account that if $K$ is compact and $F=\mathbb R\setminus U$ is disjoint and closed, then $dist(K,F)>0$. You may already have seen a proof that the distance between a compact set and a disjoint closed set is positive (e.g. this or this or this or this) then proceed as follows. Take $t>0$ such that $dist(K,\mathbb R\setminus U)>t$. Let $V=(-t,t)$. Then $K+V$ is disjoint from $\mathbb R\setminus U$, that is $K+V\subseteq U$. Then the rest of the proof of Steinhaus theorem seems correct. 
It seems that one may fix Wikipedia's proof (remaining close to their model) by taking $V=(-\frac\delta2,\frac\delta2)$ where $\delta$ is the Lebesgue number for the cover $\{k+V_k:k\in K\}$. But I think their model is the result of an unnecessary edit, and should not be followed to every detail. 
In the original January 2007 entry Wikipedia says "..Since $K\subset U$, there is an open cover of $K$ that is contained in $U$. $K$ is compact, hence one can choose a small neiborhood $V$ of $0$ s.t. $K+V\subset U$." Details were not supplied, and perhaps should not have been, though I find the first sentence above redundant and/or misleading. 
The incorrect part was introduced in July 2015 perhaps in an attempt to provide more details, unfortunately is seems this was not done carefully enough. My guess is that Karl Stromberg (to whom this proof is attributed) is not to be blamed for the additionally supplied incorrect details. 
I have personally corrected in the past a Wikipedia entry, and recently seen another post indicating that Wikipedia was not precise. You may type "wikipedia" in MSE search field, and find many similar questions,  indicating that Wikipedia may occasionally be mathematically imprecise. 
