Exercise 3, § 1

Tag 03HC, permanent. Cite it as [Bourbaki, Tag 03HC].

Theory of Sets, II, 123

Let A and B be two subsets of a set X. Show that the relation $B \subset \complement A$ is equivalent to $A \subset \complement B$ and that the relation $\complement B \subset A$ is equivalent to $\complement A \subset B$ .

Solution

This solution was written by laguna-s-2.1-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

The equivalence $B \subset \complement_{\mathrm{X}} A \Leftrightarrow A \subset \complement_{\mathrm{X}} B$ follows from the duality between subset relations and their complements stated in §1, after Definition 3 (complement of a set). Specifically, for any subsets $P$ and $Q$ of $\mathrm{X}$ , the relation $P \subset Q$ is equivalent to $\complement_{\mathrm{X}} Q \subset \complement_{\mathrm{X}} P$ . Applying this with $P = B$ and $Q = \complement_{\mathrm{X}} A$ , we have:

B \subset \complement_{\mathrm{X}} A \quad \text{if and only if} \quad \complement_{\mathrm{X}} (\complement_{\mathrm{X}} A) \subset \complement_{\mathrm{X}} B.

Since $\complement_{\mathrm{X}} (\complement_{\mathrm{X}} A) = A$ , this reduces to:

A \subset \complement_{\mathrm{X}} B.

Similarly, the relation $\complement_{\mathrm{X}} B \subset A$ is equivalent to $\complement_{\mathrm{X}} A \subset B$ by applying the same duality with $P = \complement_{\mathrm{X}} B$ and $Q = A$ . This gives:

\complement_{\mathrm{X}} B \subset A \quad \text{if and only if} \quad \complement_{\mathrm{X}} A \subset \complement_{\mathrm{X}} (\complement_{\mathrm{X}} B) = B.

Hence, both equivalences are established.

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