Infinite cardinals formal proof

Question

I need help in showing

$$2^a + 2^a = 2 · 2^ a$$

It sounds obvious but how can I formally prove it?

Answer 1

Let $A$ be a set of cardinality $a$, and let its power set be $\mathcal{P}(A)$; take also another set $Y$ which bijects with $\mathcal{P}(A)$ but is disjoint, and let $f: \mathcal{P}(A) \to Y$ be a bijection.

We wish to show that there is a bijection between the following two sets:

• $\mathcal{P}(A) \sqcup Y$ (where $\sqcup$ notes that the union is disjoint), of cardinality $|\mathcal{P}(A)| + |Y| = 2^a + 2^a$
• $2 \times \mathcal{P}(A)$ (the set of $2$-tuples such that the first element is $0$ or $1$ and the second is a subset of $A$), of cardinality $2 \times 2^a$.

There's an obvious injection from the second to the first: take $(0, x)$ (where $x \subseteq A$) and output $x$, while take $(1,x)$ and output $f(x)$ the isomorphic copy of $x$ in $Y$.

You can easily check that this is actually surjective too: it hits all of $\mathcal{P}(A) \sqcup Y$.

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If you're worried about the ability to pick the set $Y$, we can define it as $$Y = \{ x \sqcup \{ \mathcal{P}(A) \}: x \in \mathcal{P}(A) \}$$ where the union inside the set-builder is a disjoint union since $\mathcal{P}(A)$ cannot lie in $\mathcal{P}(A)$. Then $Y$ is disjoint from $\mathcal{P}(A)$ because no member of $\mathcal{P}(A)$ can contain $\mathcal{P}(A)$.