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I need to find an example of set $X$ with five elements that every one of them is subset of this set $X$. does the next set - $X=\{\{1\},\{2\},\{3\},\{4\},\{5\}\} $ answer this question?

Edit: A good example of a set with 4 elements will be the power set of $\{\varnothing,\{\varnothing\}\}$ ?

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  • $\begingroup$ Your answer doesn't work: $\{1\} \not\subseteq X$ since $1 \not\in X$. Try constructing a set of five elements, one of which is the emptyset. $\endgroup$ – leibnewtz Nov 1 '17 at 20:00
  • $\begingroup$ $\{1\}$ is not a subset of your example. It is an element of which is a different property. No, your example does not work. $\endgroup$ – JMoravitz Nov 1 '17 at 20:01
  • $\begingroup$ As a hint, the way in which natural numbers are defined according to the Von Neumann construction might be useful. $\endgroup$ – JMoravitz Nov 1 '17 at 20:04
  • $\begingroup$ @JMoravitz Please check my edit $\endgroup$ – kicklog Nov 1 '17 at 20:14
  • $\begingroup$ Hint: Try the same question with $5$ replaced by a smaller number such as $2$, $1$, or $0$: e.g., a set $X$ with one element, such that that one element is also a subset of $X$. Extra bonus hint: What is something that is always guaranteed to be a subset of $X$, in fact, a subset of every set? $\endgroup$ – Zach Teitler Nov 1 '17 at 21:00
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As mentioned twice already in the other answers, the number $5$ according to the Von Neumann construction of the natural numbers fits the criteria that you are looking for. It can be seen from the construction that for any natural number $n$ and natural number $m$ with $m<n$ that $m\in n$ as well as $m\subseteq n$. (In fact, with this construction of the natural numbers, this is how the less than relation is defined in the first place!)

A simpler construction which is easier to read which also works is this:

$$A=\{\emptyset,\{\emptyset\},\{\{\emptyset\}\},\{\{\{\emptyset\}\}\},\{\{\{\{\emptyset\}\}\}\}\}$$

That is, the set contains the empty set enclosed in no brackets, one pair of brackets, two pairs of brackets, on up to four pairs of brackets. Notice that $\emptyset$ is a subset of the above trivially since $\emptyset$ is a subset of every set. Further, the empty set enclosed in a $k$ sets of brackets (with $1\leq k\leq 4$) is a subset because $A$ has as an element the empty set enclosed in $k-1$ sets of brackets as well.

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  • $\begingroup$ for 4 elements, does the next set $\{\varnothing,\{\varnothing\},\{\{\varnothing\}\},\{\varnothing,\{\varnothing\}\}\}$ answer the question? $\endgroup$ – kicklog Nov 1 '17 at 22:47
  • $\begingroup$ @ovedpoovedpo yes, and is kind of a mix between my two answers above. $\endgroup$ – JMoravitz Nov 1 '17 at 23:24
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what about $\{ \{\},\{\{\}\},\{ \{\}, \{\{\}\} \},\{\{\}, \{\{\}\},\{ \{\}, \{\{\}\} \}\},\{ \{\}, \{\{\}\},\{ \{\}, \{\{\}\} \},\{\{\}, \{\{\}\},\{ \{\}, \{\{\}\} \}\}\}$

this thing looks sick.

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  • $\begingroup$ Can you explain what construction this is? It's a bit hard to read. $\endgroup$ – Carl Schildkraut Nov 1 '17 at 20:42
  • $\begingroup$ @CarlSchildkraut It is precisely what my hint above should have lead the OP to. It is the formal definition of the number five according to the Von Neumann construction of the natural numbers. It is quite hard to read in this form however. I would have written it in multiple lines, in effect defining all of the numbers leading up to five and using the numerical representations instead of the set representations where possible in later lines. $\endgroup$ – JMoravitz Nov 1 '17 at 21:15
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The answer above is rather awkward. What it's saying is, let $$S_0 = \emptyset$$ $$S_1 = S_0 \cup \lbrace S_0 \rbrace$$ $$S_2 = S_1 \cup \lbrace S_1 \rbrace$$ ... and so on ...

Each of the sets has all the elements of the set before it, plus one new one. Moreover, every element of each of the sets is a subset of the set.

The awkward answer with all the braces is simply expanding out $S_5$.

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