Multiple of Semiperfect Number is Semiperfect

Theorem
Let $n \in \Z_{>0}$ be a semiperfect number.

Let $k \in \Z_{>0}$ be a (strictly) positive integer.

Then $k n$ is also a semiperfect number.

Proof
Let $P$ be a subset of the divisors of $n$ such that the sum of the elements of $P$ equals $n$.

Let $\sigma = \ds \sum_{p \mathop \in P} p$ be the sum of the elements of $P$.

Let:
 * $Q = \set {k p: p \in P}$

be the set of elements of $P$ multiplied by $k$.

We have by definition that:
 * $\forall p \in P: p \divides P$

where $p \divides P$ denotes divisibility.

Then we have:
 * $\forall p \in P: k p \divides k P$

Thus $Q$ constitutes a subset of the divisors of $k n$.

But:

demonstrating that $Q$ constitutes a subset of the divisors of $k n$ equal to $k n$.

Hence the result, by definition of semiperfect number.