Let $\{A_\alpha:\alpha \in \Lambda\}$ be an indexed collection of sets. If $\bigcap \{A_\alpha:\alpha \in \Lambda\} \neq \emptyset$, then for each $\beta \in \Lambda$, $A_\beta \neq \emptyset$.

My thought was a proof through contraposition:

Assume $A_\beta = \emptyset$ for some $\beta \in \Lambda.$ It would follow that the intersection of $A_\beta$ with another set $A_\gamma$, where $\gamma \in \Lambda$ would yield the empty set. Thus $\bigcap \{A_\alpha:\alpha \in \Lambda\} = \emptyset$.

Is this proof valid or did I maybe overlook something. Any thoughts would be appreciated.

  • 4
    $\begingroup$ It is valid, but could be shorter, observing that the intersection of all the $A_\alpha$s is contained in any $A_\beta$. $\endgroup$ – Bernard Jun 6 at 20:57

Going off of Bernard's comment, recall that for any sets $A$ and $B$ we have $(A \cap B) \subset A$. Therefore, supposing by contraposition that $A_k = \emptyset$ for some $k \in \Lambda$, we see that $\bigcap_{\alpha} A_{\alpha} \subset A_{k} = \emptyset$. Thus $\bigcap_{\alpha} A_{\alpha} = \emptyset$, and the proof is complete.

(Edit: to clarify, yes, your proof is valid, though this approach is slightly more direct.)


If $\{A_\alpha|\;\alpha\in\Lambda\}$ is a non-empty family of sets, and $\bigcap\{A_\alpha|\;\alpha\in\Lambda\}\not=\varnothing$, then let $a$ be one of its elements. By definition of the intersection of a set, $a$ is such that $a\in A_\alpha$ for all $\alpha\in\Lambda$. Now, $a\not\in\varnothing$, and therefore, $A_\alpha\not=\varnothing$, for all $\alpha\in\Lambda$, by the axiom of extensionality.


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