What is the sum of two countably infinite sets? Another countably infinite set? I am asked to find this in a question.
Asked
Active
Viewed 2,436 times
2
3 Answers
7
Hint: Assuming "sum" means "union", let $A = \{a_1, a_2, \dots\}$ and $B = \{b_1 , b_2, \dots\}$ be two countable sets. Consider the map
$$ f(n) = \left\{ \begin{array}{ccc} b_{n/2} & & n \text{ is even} \\ a_{\frac{n+1}{2}} & & n \text{ is odd} \end{array} \right.$$
Show that $f(n)$ is a bijection from $\mathbb{N} \to A \cup B$.
JavaMan
- 13,153
-
3JavaMan's map allows you to count the elements of the union (thus witnessing its countability) by counting as follows: "First element of $A$, First element of $B$, Second element of $A$, Second element of $B$, ..." – Austin Mohr Nov 25 '11 at 20:14
1
For the operation suggested in the question, the proper name is 'union', not sum. The sum of two sets (the Minkowski sum) can be defined as
$$ A+B=\{a+b : a \in A, b \in B\}.$$
In this case, it can be proved that if $A,B$ are countable, then so is $A+B$. First, note that $A \times B \simeq \Bbb{N}\times \Bbb{N}\simeq \Bbb{N} $ ( $ \simeq$ means that they have the same cardinal number).
Then the function $f: A\times B \to A+B $ defined by $f(a,b)=a+b$ is surjective by definition. This means that $\operatorname{card}(A \times B) \geq \operatorname{card}(A+B)$, therefore $A+B$ is countably infinite also.
Srivatsan
- 26,311
Beni Bogosel
- 23,381
-
The sum can also be defined in cardinalities. The sum of two sets is their disjoint union (take ${0}\times A\cup{1}\times B$, for example.) – Asaf Karagila Nov 25 '11 at 21:33
@1you can - and should - write the username it will then notify that user of your reply. – Asaf Karagila Nov 25 '11 at 20:05