# Proving bijectivity of a function

Let $$X, Y$$ be sets such that $$X \subseteq Y$$ and $$\omega\in X^C$$ (complement of $$X$$), and let $$B := \{ A \uplus \{\omega\} : A \in 2^X\}$$.

Then the function $$F: 2^X\to B; A \to A \uplus \{\omega\}$$ is bijective.

I feel as if I don't understand the question at all, and if I did, I still wouldn't be able to prove this. Anyone who can help me get started?

• What is the meaning of $\uplus$ ?
– ZAF
Commented Mar 20, 2020 at 18:46
• ⊎ means a disjoint union Commented Mar 20, 2020 at 18:50
• $F: 2^{X} \to B$, isn't it ?
– ZAF
Commented Mar 20, 2020 at 18:52
• My bad! I'll edit it right now Commented Mar 20, 2020 at 18:57

If $$X\not= Y$$, then there exists $$\omega \in X^{C}$$

Let see $$F$$ is a bijective function

Let $$W \in B$$, then there exist $$A \in 2^{X}$$ (i.e. $$A \subset X$$) such that $$W = A \uplus \{ \omega \}$$

So $$W = F(A)$$, then $$F$$ is a surjective function

Now, suppose $$F(A) = F(K)$$ for some $$A,K \in 2^{X}$$

We have that $$A \uplus \{ \omega \} = K \uplus \{ \omega \}$$

If there exist $$x \in A \subset X$$ such that $$x \notin K \implies x \in \{ \omega \} \implies x = \omega \in X^{C}$$ is a contradiction

Then $$x\in A \implies x \in K$$ so $$A \subset K$$

And the same if $$x \in K \implies x \in A$$ then $$K \subset A$$

Thus $$A = K$$

So $$F$$ is injective

Then $$F$$ is a bijection