Set Theory - Subset of set

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

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

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 ;-)