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I have labelled this $$ \{\{1\}\}\subseteq\{1,2,\{1,2\}\} $$ as true, it is a subset of the third element, is this true?

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4 Answers 4

First let us recall the definition of $\subseteq$: we say that $A\subseteq B$ if and only if every $x$ which is an element of $A$ is an element of $B$, i.e. if $x\in A$ then $x\in B$.

This can be confusing when dealing with concrete sets. Try replacing them with variables:

$$a=\{1\}, b=\{a\}, c=\{1,2\}$$

Now your problem amounts to $b\subseteq c\cup\{c\}$. Since $b=\{a\}$ this is the same as asking if $a\in c\cup\{c\}$. This would be true if and only if $a\in c$ or $a=c$.

Since $a\neq c$ this again amounts to $a\in c$. However $x\in c$ if and only if $x=1$ or $x=2$, but $a$ is neither of those.

Therefore $\{\{1\}\}\nsubseteq\{1,2,\{1,2\}\}$.

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I agree with previous answers: $\{\{1\}\} \subseteq \{1,2,\{1,2\}\}$ is the same as $\{1\} \in \{1,2,\{1,2\}\}$ and this is not true, the only set inside is $\{1,2\} \neq \{1\}$.

However, if this is a set theory course, there might be some tricks given that numbers are defined as sets: $0 = \{\}, 1 = \{0\} = \{\{\}\}, 2 = \{0,1\} = \{\{\},\{\{\}\}\}$. For example normally $\{0,1\} \notin \{2\}$, but if we were very picky, then $\{\{\},\{\{\}\}\} \in \{\{\{\},\{\{\}\}\}\}$. Of course, this is madness, but to be 100% sure we should also check that $$\{1\} = \{\{\{\}\}\} \notin \{\{\{\}\},\{\{\},\{\{\}\}\},\{\{\{\}\},\{\{\},\{\{\}\}\}\}\} = \{1,2,\{1,2\}\}$$ which fortunately agrees with the first answer.

I hope I hadn't confused you too much ;-)

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The inclusion is not true, since $\{1\}$ is not an element of $\{1,2,\{1,2\}\}$.

In order for something to be a subset of a bigger set, each element of the subset must be an element of the whole set.

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Why is this? {{1}} why is that not a subset of the third element? –  JamieB Apr 22 '12 at 11:31
1  
For $\{\{1\}\}$ to be a subset of the third element $\{1,2\}$, if I understand what you mean, then it would require $\{1\}$ to be an element of $\{1,2\}$, which is not the case. And by the way, even if it would be a subset of the third element, it would not imply that it would be a subset of the whole collection.. –  Thomas E. Apr 22 '12 at 11:33
    
$\{1\}$ is the only element of the set $\{\{1\}\}$, but $\{1\}$ does not show up in the list defining the other set. –  Andrea Mori Apr 22 '12 at 11:44

Compute the powerset of $X =\lbrace 1,2,\lbrace 1,2\rbrace\rbrace$. So, we have that $P(X)=\lbrace\emptyset,X,\lbrace 1\rbrace,\lbrace2\rbrace,\lbrace\lbrace1,2\rbrace\rbrace,\lbrace 1,2\rbrace,\lbrace1,\lbrace 1,2\rbrace\rbrace,\lbrace2,\lbrace1,2\rbrace\rbrace\rbrace$. These are all of the subsets of $X$. Notice $\lbrace\lbrace1\rbrace\rbrace\not\in P(X)$, and thus $\lbrace\lbrace1\rbrace\rbrace\not\subseteq X$.

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