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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-...
Michael Chu's user avatar
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,...
M.H.Hooshmand's user avatar
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 ...
Bilbottom's user avatar
  • 2,658
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{...
user avatar
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 ...
Henry's user avatar
  • 5,719
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 ...
Bumblebee's user avatar
  • 18.4k
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 ...
shinzou's user avatar
  • 4,011