Let $A_1 \cap A_2 \cap \dots \cap A_n \neq \varnothing $. Then $A_1 \cup A_2 \cup \dots \cup A_n \neq \varnothing$.

Question

Let $A_1,A_2,\dots, A_n$ be sets such that $A_1 \cap A_2 \cap \dots \cap A_n \neq \varnothing $ holds for all $n$. Then $A_1 \cup A_2 \cup \dots \cup A_n \neq \varnothing$.

Is the following proof correct?

Proof:

If $A_1 \cap A_2 \cap \dots \cap A_n \neq \varnothing $, then $A_1,A_2,\dots, A_n$ must all have at least one common element. Therefore the sets $A_1,A_2,\dots, A_n$ are all non-empty. Hence there exists one non-empty set among $A_1,A_2,\dots, A_n$. Hence $A_1 \cup A_2 \cup \dots \cup A_n \neq \varnothing$.

Answer 1

It's the obvious thing, as, $A_1 \cap A_2 \cap \dots \cap A_n \subseteq A_1 \cup A_2 \cup \dots \cup A_n $.

Answer 2

Writing formal details: $$A_1 \cap A_2 \cap \dots \cap A_n \neq \varnothing \Rightarrow\\\Rightarrow \exists x \in A_1 \cap A_2 \cap \dots \cap A_n \Rightarrow\\\Rightarrow \exists i, 1 \leqslant i \leqslant n, x \in A_i \Rightarrow\\ \Rightarrow\ x \in A_1 \cup A_2 \cup \dots \cup A_n \Rightarrow\\ \Rightarrow A_1 \cup A_2 \cup \dots \cup A_n \neq \varnothing$$

Answer 3

Yes, the proof is correct but you are using too many not really needed conditions, at least not to that precision. You say there is at least one common element. We need far less than that. I would prove this using the opposition:

If $A_1 \cup A_2 \cup \dots \cup A_n = \varnothing$ then $A_1 = A_2 = \dots = A_n = \varnothing$ and then $A_1 \cap A_2 \cap \dots \cap A_n = \varnothing$

because no set in the union can have an element.

So if you have:

$$\exists A_1,A_2,...,A_n; A_1 \cap A_2 \cap \dots \cap A_n \neq \varnothing \Rightarrow A_1 \cup A_2 \cup \dots \cup A_n = \varnothing$$

then you would have:

$$\exists A_1,A_2,...,A_n; A_1 \cap A_2 \cap \dots \cap A_n \neq \varnothing \Rightarrow A_1 \cup A_2 \cup \dots \cup A_n = \varnothing \Rightarrow A_1 \cap A_2 \cap \dots \cap A_n = \varnothing$$

which is an obvious contradiction.