Is $1$ a subset of $\{1\}$

Question

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...

${1}$ 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 $A$. — Miguelgondu, Sep 02 '14 at 20:43
{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}. — , Sep 02 '14 at 22:16

score 21 · Accepted Answer · answered Sep 02 '14 at 20:47

21

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

answered Sep 02 '14 at 20:47

idm

11,824

3

No flaws in your logic, but that doesn't really answer the question... – Dennis Sep 02 '14 at 23:14
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@Dennis: great !! Therefore propose a real answer ! – idm Sep 02 '14 at 23:22
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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. – Dennis Sep 02 '14 at 23:28
My last comment was absolutely not a reproach. I just incited you to make an answer that would be more helpful for the OP :-) – idm Sep 02 '14 at 23:30
@idm Is {1} a subset of {{1}}? – A user Sep 08 '14 at 23:08
no, an element ! – idm Sep 09 '14 at 09:18

score 15 · Answer 2 · answered Sep 02 '14 at 20:50

15

$1$ 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).

answered Sep 02 '14 at 20:50

Asaf Karagila

393,674

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. – user26486 Sep 02 '14 at 22:51
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@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 – Dennis Sep 02 '14 at 23:05
Using this representation of numbers, we would, however, have $1\subset 2$. Weird, but true. – Dan Christensen Sep 03 '14 at 18:15
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Seeing how several people gave similar answers, the downvote feels kinda personal. – Asaf Karagila Nov 04 '14 at 03:47

Dennis · Answer 3 · 2014-09-02T23:31:29.120

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:

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

score 8 · Answer 4 · edited Sep 02 '14 at 22:11

8

$1 \in \{1\}$

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

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

edited Sep 02 '14 at 22:11

answered Sep 02 '14 at 20:49

Aldo

350

santos_mgr · Answer 5 · 2014-09-03T07:06:23.750

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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}

edited Sep 03 '14 at 07:06

answered Sep 03 '14 at 06:49

santos_mgr

31

user153012 · Answer 6 · 2014-11-02T12:38:15.247

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 $1 \in \{1\}$.

score 1 · Answer 7 · answered Sep 03 '14 at 03:56

1

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.

answered Sep 03 '14 at 03:56

Sparr

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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 – Asaf Karagila Sep 03 '14 at 03:58
I know that feeling. There are a couple of things that "working mathematicians" sometimes do that give set theorists the creeps... :) – Frunobulax Sep 03 '14 at 06:28

Is $1$ a subset of $\{1\}$

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