Proof confirmation regarding the intersection of an indexed collection of sets

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.

• It is valid, but could be shorter, observing that the intersection of all the $A_\alpha$s is contained in any $A_\beta$. – 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.
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.