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

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

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

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

where $p \mathrel \backslash P$ denotes divisibility.

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

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

But:
 * $n k = k \sigma = k \displaystyle \sum_{p \mathop \in P} p = \displaystyle \sum_{p \mathop \in P} k p = $

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

Hence the result, by definition of semiperfect number.