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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 $A_1 \cup A_2 \cup \dots \cup A_n \neq \varnothing$.

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    $\begingroup$ Sure, though it might be simpler to note that $A_1\cap B\neq \emptyset\implies A_1\neq \emptyset\implies A_1\cup C\neq \emptyset$. $\endgroup$
    – lulu
    Commented Jul 27, 2020 at 15:56
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    $\begingroup$ Having "for all $n$" in the first line of the question is confusing! If $n$ is some fixed number, you don't need that. The reason one might say something like "for all $n$" is if they are starting with an infinite collection of sets $A_1, A_2, \dots$. $\endgroup$
    – Adina Goldberg
    Commented Jul 27, 2020 at 16:04

3 Answers 3

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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 $.

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

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    $\begingroup$ This answer uses more symbols than the answer written by @hjksdk, but OP's answer is more readable and clearer. $\endgroup$
    – Adina Goldberg
    Commented Jul 27, 2020 at 16:05
  • $\begingroup$ @Adina Goldberg. Words "more readable and clearer" are very individual and totally dependent on mathematical tastes. The number of characters can be counted and used as a measure, but do you think this measure is accepted throughout the mathematical world as an estimate for the values from the first sentence? To down voting is your right, but, possibly, it is better to base it on objective criteria. $\endgroup$
    – zkutch
    Commented Jul 27, 2020 at 16:16
  • $\begingroup$ That's a fair point. I suppose this is objectively a different way to write the answer, and some may prefer it. I do not think this translation helps OP determine if they have written a correct proof or not. $\endgroup$
    – Adina Goldberg
    Commented Jul 27, 2020 at 16:32
  • $\begingroup$ Why do you have $\Rightarrow$ at the end and the start of each line? $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 27, 2020 at 17:49
  • $\begingroup$ For same reason as when we make it for "$=$" : better readability , convenient when moving from one long line to another etc. Some texts prefer to keep such symbols on first line only, some only on second(Bourbaki). For formal representation I prefer last. Some on both. On this site I already meet case, when votes starts only after I divide text on this manner. I am sure you know all these reasons, No? $\endgroup$
    – zkutch
    Commented Jul 27, 2020 at 18:14
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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.

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