All Questions
Tagged with sumset elementary-set-theory
7
questions
0
votes
1
answer
90
views
How many subsets $S$ of integer interval $[0,n]$ such that $k \not \in S+S$?
After a bit of experimentation, I thought of the following conjecture:
Given any $n \in \mathbb{Z}_{\geq 0}$ and $k \in [0,2n]$, we have $$|\{S : (S \subseteq [0,n]) \land (k \not \in S+S)\}| = 2^{|n-...
4
votes
1
answer
183
views
Two uncountable subsets of real numbers without any interval and two relations
Are there two uncountable subsets $A, B$ of real numbers such that:
(1) $(A-A)\cap (B-B)=\{ 0\}$,
(2) $(A-A)+B=\mathbb{R}$ or $(B-B)+A=\mathbb{R}$ ?
We know that if one of them contains an interval,...
6
votes
1
answer
2k
views
The sum of a set $A$ with the empty set, $\varnothing$
Given that the sum of two sets is defined as
$$
A + B = \big\{ a + b : a \in A, b \in B \big\},
$$
how might one compute the sum
$$
A + \varnothing
$$
where $A$ may or may not be empty? In his book ...
12
votes
3
answers
4k
views
If $C$ is the Cantor set, then $C+C=[0,2]$.
Question : Prove that $C+C=\{x+y\mid x,y\in C\}=[0,2]$, using the following steps:
We will show that $C\subseteq [0,2]$ and $[0,2]\subseteq C$.
a) Show that for an arbitrary $n\in\mathbb{...
1
vote
3
answers
221
views
Show that $|A+A|\geq (2n-1)$
Consider a set $A$ consisting of $n$ natural numbers $\{a_i\}_{i=1}^n$ such that $a_1<a_2 < \cdots <a_{n-1} < a_n$. Define the set $A+A$ such that it contains $a_i + a_j \ ; \ i \leq j$ as ...
3
votes
1
answer
324
views
What are the bounds (upper and lower) for $|A+A|$?
Let $A$ be a finite set of real (or complex) numbers. If I consider sets with small sizes, we have that:
If $A$ is the empty set, then $A+A$ is also empty.
If $A$ is a singleton, then $A+A$ is ...
1
vote
2
answers
65
views
Proving if $|A|\ge 4 \vee |A|\le 2$ then $|A+A|\neq 4$ with direct, contradiction and contraposition
Prove if $|A|\ge 4 \vee |A|\le 2$ then $|A+A|\neq 4$. $A$ is some set and we define $A+B=\{a+b|a\in A, b\in B\}$, $A$ is some subset of the reals.
In a direct proof and proof by contradiction I'd ...