# Problem understanding the concept of a union

$A\cup B$ means that at least one of $A$, $B$ happens and $\bigcup_{k=n}^{\infty}A_k$ means that at least one of the events $A_k$ for $k\geq n$ happens. However, I am having trouble understanding this as intuitively I see $A\cup B$ as being a set that contains elements of BOTH $A$ and $B$ as it seems like $A\cup B$ contains elements in both $A$ and $B$. However, this is the definition of an intersection. Is there a way to intuitively understand why $\bigcup_{k=n}^{\infty}A_k$ implies least one of the events $A_k$ for $k\geq n$ happens? Would someone be able to give an intuitive example where we have $\bigcup_{k=n}^{\infty}A_k$ but ONLY one $A_k$ occurs? Thanks!

• I don't understand this sentence: "intuitively I see $A\cup B$ as being a set that contains elements of BOTH $A$ and $B$ as it seems like $A\cup B$ contains elements in both $A$ and $B$." It's unclear what you think the "union" symbol means, or why you would have some kind of "intuition" about it based on something other than the definition itself. – Kyle Strand Sep 17 '15 at 22:21
• To the last part of your question: Let the $n=0$ bet the moment (in seconds) you asked this question. Let $A_k$ mean that someone has answered in second $k$. The infinity union means you have an answer. If there was only one single answer the you would have only one $A_k$. Intuitive enough? – Thinkeye Sep 18 '15 at 10:54

Contains both the elements of $A$ and $B$ (union, contains everything $A$ or $B$ contain)

Contains the elements of both $A$ and $B$ (intersection, contains only what $A$ and $B$ share)

Don't get confused with english, as it is not rigorous. use it to get an intuition of what's going on, but when it starts to be confusing, stick to the mathematical definition ;-)

• Actually I think you've demonstrated that English is sufficiently rigorous for this purpose. The questioner's mistake is that not all sentences involving the word "both" are synonymous, and his intuition has chosen the wrong one! – Steve Jessop Sep 18 '15 at 0:49

Note the difference between:

• elements $x$ of both $A$ and $B$ (i.e., $A \cap B$)
• both elements $a$ of $A$ and $b$ of $B$ (i.e., $A \cup B$)

The phrase "elements of both $A$ and $B$" is therefore ambiguous; it could mean either of the two.

Thus $\bigcup\limits_{k=n}^\infty A_k$ can be described as "the collection of elements $a_k$ of $A_k$ for each $k$".

I think that by trying to think of sets as events, you are doing yourself a disservice and muddying your own understanding. Instead, just think of sets formally for the moment, as collections of elements with no built-in interpretation or meaning.

$A \cup B$ is by definition $\{x : x \in A \mbox{ or } x \in B\}$.

By way of explanation, I would ideally like to stop there -- there is no interpretation that can clarify this simple operator. Similarly,

$\bigcup_{\alpha \in S} A_\alpha := \{x : x \in A_\alpha \mbox{ for some } \alpha \in S\}.$

Typically, trying to understand definitions like this by coming up with different interpretations is counterproductive. That is not always necessarily true, but I think it may be the case here. In plain English, though, $A \cup B$ is the set of all elements in either $A$ or $B$, and $A \cap B$ is the set of all elements in both $A$ and $B$, and you will rarely be led astray by thinking of these sets with their Venn diagram equivalents (at least early on in your studies of math):

Since you seem to be studying probability, pictures like this are helpful to remind yourself of Bayes' Theorem (at least, they were helpful to me).

Intuitive examples are tricky, because they depend on the intuition of the individual. However, it is worth trying, since you just ask for an example (which is often easier than finding an intuition for the whole thing)

Let's say we're looking for some defect in a part. We can label each of the defect with lower case letters: q, r, s, t, u, v, w, x, y, z. My, my, that's a lot of different flaws that the part could have. Our quality control people would have to be well trained individuals to pick up any one of those defects, and any one defect is unacceptable to our customers!

So, naturally, we divide the work up. We could divide it into three worker's worth of work. We can label the workers with capitol letters, each of which is looking for sets of letters:

A = {r, s, t}
B = {u, v, w}
C = {x, y, z}

Now, we have three workers who all want to look for one of three defects, and they'll raise their hand if they see anything. Thus, we can say that the set of defects we can detect is $A \cup B \cup C$, that is, "the union of A, B, and C."

Now there's a disadvantage to this, which is that each defect is only looked at by one worker. Let's add a little workload, and make sure that each defect is checked by two people:

A' = {r, s, t, u, x}
B' = {u, v, w, s, y}
C' = {x, y, z, t, w}

Now, the set of defects that we can detect is $A^\prime \cup B^\prime \cup C^\prime$, or "the union of sets A', B', and C'"

Now if anyone asks if that means we are being more stringent in our testing, catching more types of defects, we say "no" because $A \cup B \cup C = A^\prime \cup B^\prime \cup C^\prime$ We're testing the same set of defects, we're just doubling up on how we go about looking for them.

Now, as for they symbols themselves... I'm a visual person when it comes to remembering which is which. I feel sorry for subjecting you to my mnemonic, but here's what I use. I remember $\cup$ as a cup that's held open on the top, trying to catch everything it can (so it collects everything that is in either set or in both sets), while $\cap$ is a cup you're putting down into the middle of the sets, like a cup to capture a spider (so it collects everything that is in both sets, but nothing that is only in either). You can try to capture as much rain as you want, by moving the $\cup$ back and forth to try to capture all the raindrops, but the $\cap$ can only capture one spider, even if there's two spiders in the room. I use a related mnemonic for the logical symbols: $\lor$ (OR) tries to capture either event as they fall downward, while $\land$ (AND) wont trigger if one or the other is true (it lets them slide down one side of the angle or the other), but both are true, it lands in the middle and doesn't slide off.

Silly mnemonics perhaps, but pretty much all mnemonics are!

As the other answers indicate, one must be careful about interpreting union and intersection using the informal words "and" and "or". Observe that when we say, "your choices include chicken and steak", that means the same thing as "you can have either chicken or steak", not "you must have chicken and steak".

Suppose you flip a fair coin. There are two possible outcomes: heads ($H$) and tails ($T$). The set of events is $\{ \emptyset, \{H\}, \{T\}, \{H, T\}\}$. The "meaning" of these four events is "neither heads nor tails", "heads", "tails", and "either heads or tails". These four events have associated probabilities $0, 1/2, 1/2,$ and $1$.

Each event "happens" when the outcome is an element of that event; that is to say, if the coin flips heads, then both $\{H\}$ and $\{H, T\}$ "happen" (although perhaps "happen" isn't the best word for this situation). Note, in particular, that the event set $\{H, T\}$ does not mean "both heads and tails", which would of course be an impossibility for a single coin.

I think your problem is that you believe A and B have to intersect in order to describe a union. They may intersect but they don't have to.

Here's a visual.

You may be suffering from an English trap: $A \cup B$ is comprised both from elements of $A$ and from elements of $B$, $A \cap B$ is comprised of elements that are both in $A$ and in $B$. If English is tripping you up, then stop using it and revert to symbols: $x \in (A \cup B) \Leftrightarrow (x \in A) \vee (x \in B)$ or something in-between like "x is in $A \cup B$ if and only if $x \in A$ or $x \in B$.

$\bigcup_k A_k$ is comprised of elements from $A_n$, elements from $A_{n+1}$, and so forth. That is, $x$ is in $\bigcup_k A_k$ if $x \in A_n$ or $x \in A_{n+1}$ or $x \in A_{n+2}$ or .... (and cue mathematical trick to convert this infinitely-long sentence into a finite one).

Let $n=0$ and $A_k = [k, k+1)$. Then $\bigcup_k A_k$ is the set of all nonnegative real numbers.