Is it possible to multiply a set by a natural number?

Question

Say I have a set $S=\{1, 4, 10, 7\}$. Could I then multiply $S$ by $3$? Would that then look like $3S=\{3, 12, 30, 21\}$? Any help would be really appreciated.

Answer 1

Yes..you have already defined the operation..a scalar multiplication on a set.

Answer 2

Sure, we sometimes for example denote the set of even integers by $2\Bbb Z=\{\dots,-4,-2,0,2,4,\dots\}$, while the set of integers is $\Bbb Z=\{\dots,-2,-1,0,1,2,\dots\}$.

Answer 3

In fact, this is closely related to images of functions:

Let $f: A \to B$ be a function between two sets and $A' \subseteq A$, then we define $$f[A'] := \{f(a)\ |\ a \in A'\}\quad \left(= \{b \in B\ |\ \exists a' \in A'. f(a') = b\}\right)$$

Your idea is a specific instance:

Let $A' = \{1, 4, 10, 7\} \subseteq \mathbb{N} =: A$ and $f: \mathbb{N} \to \mathbb{N}$ be the operation "multiply by three", then $$f[A'] = \{3, 12, 30, 21\}$$

The other answer hints at the notation $NA'$, e.g. $3A'$.

Note that in general only $|A'| \geq |f[A']|$ holds, e.g. consider $0\mathbb{Z} = \{0\}$. However, with this "function framework", it is easy to state a criterion for when the equality holds:

If $f$ is injective on $A'$, then $|A'| = |f[A']|$.

Or in words: "If you map different elements in $A'$ to different elements in $B$, then certainly, we cannot lose elements due to duplicates in the set $f[A']$."

Can you see why $|2\mathbb{Z}| = |\mathbb{Z}|$? This proves "there are as many even numbers as integers". (Of course, you have to first define the cardinality function $|\cdot|$ for arbitrary infinite sets first.)