12
$\begingroup$

Is the number $1$ a subset of the set $\{1\}$ just as $\{1\}$ is a subset of the set $\{\{1\}\}$? I'm a little bit confused because $1$ is an element not a set...

$\endgroup$
  • 9
    $\begingroup$ $\{1\}$ is also an element of $\{\{1\}\}$, not a subset. Suppose you have $A = \{1,2,3\}$, we say $2$ is an element of $A$ (and we write $2 \in A$), but we say $\{2\}$ is a subset of $A$ (and we write $\{2\} \subseteq A$), because every element of $\{2\}$ is in $A$. $\endgroup$ – Miguelgondu Sep 2 '14 at 20:43
  • 8
    $\begingroup$ {1} isn't a subset of {{1}}, unless you are working in an ill-founded set theory and have defined 1 in a way such that 1 = {1}. $\endgroup$ – user14972 Sep 2 '14 at 22:16
  • $\begingroup$ see here, and here, and here. $\endgroup$ – user 1 Dec 31 '15 at 15:25
22
$\begingroup$

$1$ is an element of $\{1\}$, $\{1\}$ is a subset of $\{1\}$, $\{1\}$ is an element of $\{\{1\}\}$ and $\{\{1\}\}$ is a subset of $\{\{1\}\}$.

$\endgroup$
  • 3
    $\begingroup$ No flaws in your logic, but that doesn't really answer the question... $\endgroup$ – Dennis Sep 2 '14 at 23:14
  • 1
    $\begingroup$ @Dennis: great !! Therefore propose a real answer ! $\endgroup$ – idm Sep 2 '14 at 23:22
  • 3
    $\begingroup$ I didn't mean this is a bad way. It's just not immediately obvious why the fact that $1$ is an element of $\{1\}$ would imply that it cannot be a subset. $\endgroup$ – Dennis Sep 2 '14 at 23:28
  • $\begingroup$ My last comment was absolutely not a reproach. I just incited you to make an answer that would be more helpful for the OP :-) $\endgroup$ – idm Sep 2 '14 at 23:30
  • $\begingroup$ @idm Is {1} a subset of {{1}}? $\endgroup$ – A user Sep 8 '14 at 23:08
16
$\begingroup$

$1$ is generally not a subset of $\{1\}$, since $1$ is a natural number (or a real number, or whatever) and not a set. These objects are of two different types.

 

But there is something to be said here. We can represent numbers using sets. We can declare that $0$ is $\varnothing$, and that $1=\{0\}$ or $\{\varnothing\}$, and that $2=\{0,1\}$ and so on. Then a number is a set.

Still that doesn't mean that $1\subseteq\{1\}$. This would very much depend on the representation of $1$ as a set.

So while "working mathematics" is typed (meaning the type of objects matters), we can also work in an untyped environment, where everything has the same type (for example, everything is a set).

$\endgroup$
  • $\begingroup$ It looks very unusual how a set could actually be named by a single number that way, so I'd like to see some sources showing someone actually using such notation. $\endgroup$ – user26486 Sep 2 '14 at 22:51
  • 1
    $\begingroup$ @mathh: That's the usual way of defining natural number in axiomatic set theory, since the resulting set is the one provided by the Axiom of Infinity. See Set-theoretic definition of natural numbers $\endgroup$ – Dennis Sep 2 '14 at 23:05
  • $\begingroup$ Using this representation of numbers, we would, however, have $1\subset 2$. Weird, but true. $\endgroup$ – Dan Christensen Sep 3 '14 at 18:15
  • 2
    $\begingroup$ Seeing how several people gave similar answers, the downvote feels kinda personal. $\endgroup$ – Asaf Karagila Nov 4 '14 at 3:47
8
$\begingroup$

$1 \in \{1\}$

$\{1\} \subseteq \{1\}$

As you said, 1 is an element of $\{1\}$, not a set.

$\endgroup$
8
$\begingroup$

As @AsafKaragila already said, you can define natural numbers as sets. In fact, in most axiomatic set theories, that's the only way of defining numbers, since every element of a set is a set.

As for your specific question, suppose that $1\subset\{1\}$. This leaves us with two options:

  1. $1=\emptyset$ and the inclusion trivially holds.

  2. $1\neq\emptyset$. Since $\{1\}$ has only two subsets, $1=\{1\}$.

Option 1 is possible, since identifying $1$ with the empty set is perfectly valid, but it's much more natural to identify $0$ with the empty set.

Option 2 just "looks wrong", but in elementary set theory, we can't really make any statements beyond that. In any axiomatic set theory that includes the Axiom of Regularity, $1=\{1\}$ is false since $x\notin x$ for every set $x$.

$\endgroup$
3
$\begingroup$

On both the cases, 1 is element of {1} & {1} is element of {{1}}. i.e. 1 ∈{1} & {1}∈{{1}}

And, in case of subset, {1} is subset of {1}. i.e. {1}⊆{1}

$\endgroup$
3
$\begingroup$

No. Every natural number is constructed based on set theory. Because of that every natural number is a set. To be more general, because we build up the whole mathematics from set theory, we can say that in mathematics everything is a set. The construcion is the following.

  • $0 = \emptyset$
  • $1 = \{ 0 \} = \{ \emptyset \}$
  • $2 = \{0, 1 \} = \{ 0, \{0\} \} = \{ \emptyset, \{ \emptyset \} \}$
  • and so on.

Since $1 = \{ 0 \} = \{ \emptyset \}$ and $\{1\} = \{ \{ 0 \} \} = \{ \{ \emptyset \} \}$ and the subsets of $\{ 1 \}$ are $\wp\left( \{ 1 \}\right) = \wp\left(\{ \{ \emptyset \} \}\right)=\{\emptyset,\{ \{ \emptyset \} \}\}$ we can say that $ \{ \emptyset \} \notin \wp\left(\{1\}\right) $ so $\{ \emptyset \} \not\subseteq \{ \{ \emptyset \} \}$ and that is why $1 \not\subseteq \{1\}$.

On the other hand, because we have an axiom of extensionality and since we have element predicate, the fact that $\{ 1 \}$ contains $1$ exactly means that $1 \in \{1\}$.

$\endgroup$
1
$\begingroup$

It is ambiguous but not uncommon to see a set with a single member referred to as simply that member, in contexts where the author would expect the reader to understand that only sets are under discussion.

$\endgroup$
  • $\begingroup$ An unfortunate abuse of notation. I tried to have my alg. top. professor to stop doing it, but it didn't work. I kept having this twitch whenever it happened... :P $\endgroup$ – Asaf Karagila Sep 3 '14 at 3:58
  • $\begingroup$ I know that feeling. There are a couple of things that "working mathematicians" sometimes do that give set theorists the creeps... :) $\endgroup$ – Frunobulax Sep 3 '14 at 6:28

Not the answer you're looking for? Browse other questions tagged or ask your own question.