Assume $A$, $B$ are included in a common superset $D$. Verify the identity $A^c \subseteq B^c \iff B^c \subseteq A^c$

Incomplete Proof: Suppose $A^c \subseteq B^c$ and pick $x \in B^c$. We need to show $x \in A^c$

This is a problem I'm really stuck on. I'm not sure how to proceed with a direct proof. If I try to argue by contradiction and suppose that $x \in A$ I still get nowehere and if I try and use the contrpositive and show that $B^c \not\subseteq A^c \implies A^c \not\subseteq B^c$, then there'd exist an $x \in B^c$ for which $x \in A$, but again I feel like I don't have enough information to show the existence of a $y \in A^c$ such that $y \not\in B^c$

How could I go about proving this?

  • 2
    $\begingroup$ are you sure you wrote down the question correctly? $\endgroup$ – Siong Thye Goh Mar 30 '18 at 1:54
  • $\begingroup$ It's probably $A \subseteq B \Leftrightarrow B^C \subseteq A^C$ $\endgroup$ – Bram28 Mar 30 '18 at 2:04

Let $D=\{1,2,3\}$, if $A=\{1,2\}$, $B=\{2\}$, $$D \setminus A=\{3\},$$$$ D \setminus B = \{1,3\} $$

and we can see that the claim is false.

The right statement is $$A^c \subseteq B^C \iff B \subseteq A$$

Hint of one of the direction:

If $A^c \subseteq B^c$ and $x \in B$, suppose on the contrary that we assume $x \notin A$, try to get a contradiction.


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