# Can elements in a set be duplicated?

If $A = \{x \mid x \text{ is a letter of the word 'contrast'}\}$

Represent it in a Venn Diagram, and then find the $n(A)$.

Do I need to write the letter 't' twice inside the venn diagram? What should be the answer for $n(A)$?

• Sets does not have repeated elements. n(A) equals 7. Oct 29, 2012 at 13:14
• Sep 18, 2017 at 4:53

From wikipedia:

Every element of a set must be unique; no two members may be identical. A multiset is a generalized concept of a set that relaxes this criterion.

• Note that this definition has been removed from the Wikipedia page with this edit. From what I've read (I'm no mathematician), a set can have identical elements within it when it is defined, but when working with sets or comparing sets, the duplicates are ignored. So {1, 1, 2} is the same as {1, 2} is the same as {2, 1}. A multiset is where multiple instances of an element matters. Sep 7, 2021 at 18:45

No. Sets are collections where repetition and order are ignored.

Therefore in the word contrast we have the letters $\{\sf c,o,n,t,r,a,s\}$ and a simple observation tells us that there are exactly $7$ letters in this set.

• what about n(A)? Oct 29, 2012 at 10:09
• I don't know what that is. Is that "the number of elements in $A$"? Oct 29, 2012 at 10:10
• yes.the number of elements in A. Should it be 7 or 8? Oct 29, 2012 at 10:12
• @Fazlan: It’s $7$. Oct 29, 2012 at 10:17
• Is it the case that $\{a, a, b \}$ is the same in every respect as ${a, b}$ and we just take the shorter one by convention, or is it the case that ${a, a, b}$ is not a set?
– Ovi
Jun 29, 2016 at 23:23

The original question is simply ill-posed, since "letter" is ambiguous between letter-tokens and letter-types.

I'm doing the crossword. I ask you "What's an a eight letter word beginning with 'c' that means disparity?" Obviously here I mean to count letter-tokens. "Contrast" is an eight letter word in this crossword sense, as it contains eight letter-tokens though these happen to exemplify only seven different types of letter.

I challenge you: "Only use the first fifteen letters of the alphabet in your tweets today". Obviously here I'm talking about letter types. It is ok for you to tweet "Eeek. Killing me!" even though that contains many tokens of the same letter type.

Likewise talk about words is ambiguous between word-tokens or word-types. "How many words are there in Shakespeare?" could be asking how many word-tokens there are in the corpus, or how many word-types are there exemplified. (I think it is to C.S. Peirce we owe the type/token terminology for making the obvious distinction.)

• So you're saying that $A = \{x \mid x \text{ is a letter of the word 'contrast'}\}$ is ambiguous between $A = \{\text{'a'}, \text{'c'}, \text{'n'}, \text{'o'}, \text{'r'}, \text{'s'}, \text{'t'}\}$ and something along the lines of $A = \{\left(\text{'c'}, 1\right), \left(\text{'o'}, 2\right), \left(\text{'n'}, 3\right), \left(\text{'t'}, 4\right), \left(\text{'r'}, 5\right), \left(\text{'a'}, 6\right), \left(\text{'s'}, 7\right), \left(\text{'t'}, 8\right)\}$ (where a "letter" has both a type and a position)? Oct 29, 2012 at 14:56

In formal set theory, each element counts as one even though they may appear more than once, so that $\{1\}=\{1,1\}$. This is because if we let $A=\{1\}$ and $B=\{1,1\}$, then $x\in A \iff x\in B$, so that the two sets are equal by the very definition of equality of sets.

However, in elementary school texts treating sets in a more informal manner, the elements of a set are often taken to mean something else. For example, for the word contrast, its letters may be taken to be $c,o,n,t,r,a,s,t$ where there is the implicit supposition that one is dealing with $c,o,n,t_1,r,a,s,t_2$, so that the set actually has $8$ elements instead of $7$.

It all depends on the context and intention of the text.

There is nothing here that requires any double counting. You simply take the alphabet, look at the letters one by one, and decide "Does this letter appear in the word 'contrast'?" If it does, then it's in $A$, if not, then it isn't.

At least that's how I interpret the question.