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Consider that we have a set whose elements are the vowels of the English alphabet - "a", "e", "i", "o", "u". Here, "a" denotes the letter a.

In a book that I'm reading right now, this set is denoted as follows:

$$\{x | x \text{ is a vowel in the English Alphabet}\}$$

My question is what exactly $"x"$ is denoting here?

Does it denote, a single arbitrary letter of the alphabet, a single arbitrary vowel, all letters, all vowels or something else?

Mauro ALLEGRANZA
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Harshit Rajput
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    It is a variable that is ranging over all letters of the English Alphabet. The set-builder notation "select" all values of the variables that fit with the "specification": in this case all those letters of... that are vowels. – Mauro ALLEGRANZA Feb 15 '23 at 14:34
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    So if $\text {English Alphabet} = { a,b,c,\ldots, z }$ then the above set will be ${ a,e,i,o,u }$ – Mauro ALLEGRANZA Feb 15 '23 at 14:35
  • It is only a formula that translate "the set of all those letters of the English Alphabet that are vowels". – Mauro ALLEGRANZA Feb 15 '23 at 14:45
  • @MauroALLEGRANZA So if I wanted to tell about "x", before denoting this set in set-builder form, I should write - "x" denotes any (not particular) letter of the English Alphabet? So, "x" could denote "a", "b", "c", "d"... By writing first "x" we're communicating the fact that this set has letters of the alphabet, and then by writing "x is a vowel" we are communicating that we're not having all the letter but rather some specific. If I'm correct with this, I have a question that follows. – Harshit Rajput Feb 15 '23 at 14:56
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    Yes; the set-builder notation can be used to "select" a subset from an already existing set. It is a sort of procedure: "Let $x$ be a letter of the English Alphabet; if it is a vowel, then put it into the basket. If not, skip it a consider a new one." – Mauro ALLEGRANZA Feb 15 '23 at 15:00
  • @MauroALLEGRANZA Oh wow, that does clear it up to a great extent. Even though now I'm wondering what exactly a variable is. I must read a bit about it and then try to think about what you said. I would return if I still face any issues. Thank You :) – Harshit Rajput Feb 15 '23 at 15:27
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    It doesn't say anywhere what $x$ is ranging over, and with that definition, any set that includes the vowels of the English alphabet would work. Without moving too far from the subject, you could e.g. let $x$ range over the letters of the Danish alphabet (we consider "y" to be a vowel, and we have the three extra letters "æ", "ø" and "å" - that we also consider to be vowels) – Henrik supports the community Feb 15 '23 at 15:27
  • A variable is an expression used in mathematics; what we mean when we say that $(x-1)(x+1)=x^2-1$? That whatever value (number) we plug-in into the expression and we made the computation, the two sides of the identity will give the same result. – Mauro ALLEGRANZA Feb 15 '23 at 15:30
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    Two main usage: 1) to express "generality": "let $x$ be a natural number" means that the following argument will holds for every natural number. – Mauro ALLEGRANZA Feb 15 '23 at 15:31
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    And 2) to "give place" for an input value in some computation: the formula specifying a function $f$ usually is written as $f(x)= \text { formula}$ where the formula is a symbolic expression containing $x$ that gives us a sort od recipe for computing the output value (the value of the function) for every input value (the value assigned to $x$). – Mauro ALLEGRANZA Feb 15 '23 at 15:33
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    @HarshitRajput I'd say that the x in your question is like the p in the statement "let $p$ be a prime number": a parameter, a representative placeholder whose value is unspecified and which doesn't need to be instantiated, a general specification. $\quad$ Now, is the indefinite integral $x+C$ an arbitrary, representative single antiderivative, or is it a family/set of antiderivatives? – ryang Feb 15 '23 at 15:44
  • @MauroALLEGRANZA How would we represent a set of all letters of the English alphabet, in set-builder form, given that x denotes an arbitrary letter? Like this - ${x| x \text{ is a letter}}$? – Harshit Rajput Feb 17 '23 at 09:23

1 Answers1

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The $x$ used in the formula above is a variable that is ranging over all letters of the English Alphabet.

The set-builder notation selects all values of the variables that fit with the "specification": in this case all those letters of the English alphabet that are vowels.

So if the English alphabet is the set $\{ a,b,c,…,z \}$, then the above set will be: $\{ a,e,i,o,u \}$. We read it as: "the set of all those letters of the English Alphabet that are vowels".

The set-builder notation can be used in two ways: either 1) $\{ a,b,c,…,z \}$. In this case we list all elements: of course, "it works" if the number of elements is finite (and not huge).

Or 2) to "carve out" the set from another already existing set using a formula that specify the criteria to be used for selecting the elements: $\{ x \mid \text { formula with } x \}$

Thus, if we start with the set $\text {EL}$ of all the letters of English alphabet, we can use the notation to extract from it the set of vowels: $\{ x \mid x \in \text { EL and} x \text { is a vowel} \}$.

Mauro ALLEGRANZA
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