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I am a little confused about the basic definition of inclusion.

I understand that, for example, $\{4\}\subset\{4\}$.

I also understand that $4\in\{4\}$, and that it is false to say that $\{4\}\in\{4\}$.

However, is it possible to say that $4\subset\{4\}$?

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$4=\lbrace{0,1,2,3\rbrace}$ –  user91500 Jan 10 at 15:10

5 Answers 5

up vote 4 down vote accepted

First of all, sets can be elements of other sets too. For example if $X$ is a set then $\mathcal P(X)$ is the power set of $X$, and it is a set whose elements are all sets. But now that's out of the way, let us focus on the question whether or not $4\subseteq\{4\}$ makes sense.

It is possible if you interpret $4$ as a set. In naive set theory we often work under assumptions closer to type theory. There are real numbers, and there are vectors, and functions, and there are sets and there are other sort of type of mathematical objects.

In modern set theory we often work under the assumption that everything is a set. We construct surrogate sets to interpret other concepts such as the integers, or the real numbers, as sets.

For example, one of the mainstream ways to interpret the ordered pair $\langle x,y\rangle$ is by considering the set $\{\{x\},\{x,y\}\}$. Even though ordered pairs are "not sets", we can represent them using sets.

Similarly for integers, we can represent them as sets too. Often we choose the following encoding, $0=\varnothing$ and $n+1=n\cup\{n\}=\{0,\ldots,n\}$. In that case $4=\{0,1,2,3\}$. Clearly under this interpretation $4\nsubseteq\{4\}$. But under this interpretation, $0\subseteq\{0\}=1$.

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So Asaf, under that interpretation of integers as sets it is possible to say that 3 ⊂ {4}? –  Ken McGill Jan 10 at 15:19
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@Ken: Recall that $A\subseteq B$ if and only if every element of $A$ is an element of $B$. The set $\{4\}$ has one element, and $3$ has three elements (and $4$ has four). So it's fairly easy to see that the answer is negative. –  Asaf Karagila Jan 10 at 15:20
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But (to continue the investigation) $3\subset 4$, since $\{0,1,2\}\subseteq\{0,1,2,3\}$. –  rschwieb Jan 10 at 15:21

Technically it depends on the definition of 4 and the axioms of set theory you are using. With the standard definitions it is false. (Note though that $4 \subset \{4\}$ is a valid statement)

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What do you mean by "valid statement"? "Valid"where and how? –  DonAntonio Jan 10 at 15:13
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@DonAntonio: I don't think "valid" is a good word to use here (because "logically valid" is in use as a technical term meaning "will always be true", so it is confusing). What dani_s means is that "$4\subset \{4\}$" is a well-formed claim when we're working in a world where everything is sets. In ordinary everyday mathematics where we stick to an informal type system, however, $4\subset \{4\}$ is not a useful statement. –  Henning Makholm Jan 10 at 15:15
    
Exactly, that's what I meant to say :) –  dani_s Jan 10 at 15:17
    
Thanks @HenningMakholm . I thought it could be something like that dani_s meant. –  DonAntonio Jan 10 at 15:21

There are two symbols, here, and I think you may be getting them confused. I would suggest to not use the word "inclusion" (at least, not all the time) because that a different meaning in English than in math.

The $\in$ symbol is used to designate if something is inside of a set. That is, $4\in\{4\}$, and $\{4\}\in\{\{4\}\}$, but $\{4\}\not\in\{4\}$.

The $\subset$ symbol is used to show whether the elements of one set are inside another set. That is, if $A \subset B$, then $a\in B$ for every $a \in A$. Another way of looking at it: $\subset$ always has a set on both sides.

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$\;4\;$ is an element of the set $\;\{4\}\;$, so the symbol $\;\subset\;$ doesn't fit it. You need the internationally accepted symbol of curly parentheses {} or any other accepted notation in order to make clear it is a set.

For example, if $\;X = \{ 1,\{1\}\}\;$ , then we both have $\;1\in X\,,\,\{1\}\in X\;$ , and we also have $\;\{1\}\subset X\,,\,\{\{1\}\}\subset X\;$

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No, because $4$ is not a set, but an element.

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In set theory, "elements" aren't primitives, they are just other sets that have the membership relation with another set. In other words, one could say that sets and their elements are on equal footing. And very concretely, $4$ is built up as a set in elementary set theory, so this doesn't really get at the heart of the matter. –  rschwieb Jan 10 at 15:18
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Hmmm...@rschwieb : under the usual, basic and most simple rules of set theory I think Alex's answer is correct, though perhaps not thorough (but then, why should it be thorough?!) The added explanations, as in Asaf's or dani_s's, are in place but, perhaps, for a beginner could be too much to swallow in a first reading. –  DonAntonio Jan 10 at 15:25
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@DonAntonio If he meant it (and edited it) to be "$4$ is not a subset of $\{4\}$, but it is an element of $\{4\}$" then I would agree. But I think flatly declaring that a particular symbol "is not a set, but is an element" is planting a misconception. In basic set theory, there isn't such a thing as an element that isn't a set. –  rschwieb Jan 10 at 15:28
    
And by all means, someone please let me know if that edit occurs :) –  rschwieb Jan 10 at 15:34

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