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?

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

1 Answer 1


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


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