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Does any finite set $S = \{a, b, ..., c\}$ equal the union of it's elements? I believe so, but would I write that like

$$S = \{a\} \cup \{b\} \cup \cdots \cup \{c\} \tag{A}$$ $$\text{vs} $$ $$S = a \,\cup \,b \cup \cdots \cup c \tag{B}$$

Because I'm not sure how that would work out when using, say, the empty set

$$\{\} = \{\{\}\} \cup \{\{\}\} \cup \cdots \cup \{\{\}\} \tag{A}$$ $$ \text{vs} $$ $$\{\} = \{\} \,\cup \,\{\} \cup \cdots \cup \{\} \tag{B}$$

Here, the second one looks correct, although I'm curious if there's a distinction.

I was inspired to ask that while was looking at some intro probability text in which I realized an event is simply the union of the elements—although the text doesn't explicitly say that. Which is funny because the other day I asked a question on here: what is the probability of flipping two fair coins (heads and tails is $H$ and $T$, respectively) and and getting $HH$ given that we know one of the coins is $H$? Well, my previous post lead to some discussion about ambiguously worded probability problems. That specific question could be interpreted at least a few linguistically natural ways, and the conclusion was that we should avoid posing such questions that are ambiguous.

Anyway, I've been thinking about the ambiguity of such questions. In the case of the question above, the author gave the answer of the probability being $1/3$, which aligns with a natural interpretation of the question: the two coins land, we cannot see them, we only know that one of them is heads. That eliminates $TT$ from our remaining possible event set, and $HH$ is $1$ of the three events remaining. I have no disagreement about that. But, rereading the section today I came to what I hope is a legitimate realization: every event is the elements that satisfy that event connected with OR statements, i.e. the union of those elements.

Consider the above question posed slightly differently: we imagine the same scenario (and interpretation) but now we wonder what is probability of getting a heads and a tails? This question can be more precisely (and boringly) posed as: what is the probability that $HT$ OR $TH$ occurs given that there is one $H$ in the event? Here, it's clear the answer is 2/3. Notice that the only challenging part of this question was the line "...what is probability of getting a heads and a tails?". This has to (1) be translated in our minds to an appropriate subset (event) of the sample space, and at the same time we have to account for the $HT \neq TH$ relative to our sample space but that $HT$ = $TH$, relative to the event. We consider $HT$ and $TH$ unique in the sample space but the same for the event.

Sorry if that seemed superfluous. What I'm asking is: am I correct to view an event as the elements that comprise it—their union? Second, I'm curious on the notation at the top, and third, I'm curious to any methods of thinking you might have on these (or more general) probability questions that are useful in avoiding confusion. Thank you.

Ethan Bolker
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Zduff
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    To answer your question in the title, ${}$ does not equal ${{}}\cup\dots$ since on the left you have a zero element set while on the right you have a one element set. Note that ${}$ is not the same thing as ${{}}$. – JMoravitz Feb 09 '19 at 02:45
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    $S={a,b,c}={a}\cup{b}\cup{c}$ does *NOT* mean that $S$ is "the union of its elements". The sets ${a}$, ${b}$, and ${c}$ are not the elements of $S$. They are the one-element subsets of $S$. Every set (finite or infinite, empty or nonempty) is the union of its one-element subsets. – bof Feb 09 '19 at 02:47
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    As for the question relating to events in a probability space, what you were intending to remark on is that an event is the union of the atomic events which make it up, which as mentioned is technically different than being the union of the actual outcomes themselves. $a$ is different than ${a}$. – JMoravitz Feb 09 '19 at 02:50
  • Perhaps useful for the first part of your question, about notation: https://math.stackexchange.com/questions/2620616/what-is-the-difference-between-x-and-x-when-x-itself-is-a-set/2620621#2620621 – Ethan Bolker Feb 09 '19 at 03:12
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Hint: Consider again

\begin{align} \{\}&=\{\{\}\}\cup\dots\cup\{\{\}\} \\ &\stackrel{!}{=}\color{red}{\{\{\}\}}\tag{1} \end{align}

We have $$\{\}\notin\{\}\tag{2}$$ but $\{\}\in\{\{\}\}$. (Why? Revise $(2)$.) Hence equation $(1)$ (implied by transitivity of equality) does not hold.

NB: For the $\stackrel{!}{=}$, I have used the lemma that $$\bigcup_{i\in\mathcal{I}}A=A$$ for any nonempty index set $\mathcal{I}$ and any set $A$. Can you see why this lemma is true?

Shaun
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