# Proving set theory true or false

I have $A_1 \subset \bigcup_{i=1}^{n} A_i$ and need help proving if it is true or not. I believe that statement means this: $A_1 \subset (A_1 \cup A_2 \cup ... \cup A_n)$.

I believe this to be false because if every set is empty then they are all equal and it won't fit the definition of a proper subset.

Am I correct in thinking this? Or can I not set them all equal to the empty set?

• The empty set is a subset of itself. – miradulo Sep 21 '15 at 21:54
• I don't understand what you mean. – neby Sep 21 '15 at 21:55
• Many mathematicians use $\subset$ and $\subseteq$ interchangeably; that seems to be happening here. – Marcus M Sep 21 '15 at 21:57
• It all depends if your textbook uses $\subset$ for denoting proper containment. In this case the statement is false: take $n=1$ and you're done. Otherwise (and several authors use $\subset$ for denoting inclusion also in non strict sense), the statement is true. – egreg Sep 21 '15 at 21:58
• Thanks! I meant proper subset with that notation. – neby Sep 21 '15 at 22:02

If you mean for $\subset$ to be synonymouse with $\subsetneq$, then your reasoning is correct. As a counterexample, we can note that with $A_1 = A_2 = \cdots = A_n = \{1\}$, we do not have $A_1 \subsetneq \bigcup_{i=1}^n A_i$, as you stated.
There is a bit of ambiguity in the mathematical notation for subsets and proper subsets. The sign $\subseteq$ always means subset, while $\subsetneq$ always means proper subset. However, the sign $\subset$ can mean either subset or proper subset.
Usually, the notation $\subset$ stands for an ordinary subset. I believe this is because a typical mathematician will mostly encounter cases where it doesn't matter if a subset is proper or not. The notation $\subset$ is simply shorter and faster to write than $\subseteq$, so mathematicians use it for the case they work with the most.
• Oh, he meant proper subset. It was an assignment handed to us by our professor. He uses $\subset$ to denote proper subset. Thanks again! – neby Sep 22 '15 at 21:47