If $$A=\{1,{\{\emptyset}\}\}$$
then is $$B={\{\emptyset}\} \subseteq A?$$
I would say yes it is because it is the set containing the empty set, but the solution says it is not. Why is this?
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$\begingroup$ Empty set is an element of A so B is an element of A. $\endgroup$ – Yunus Syed Oct 5 '15 at 23:18
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$\begingroup$ Hi that doesnt really help to clarify for me $\endgroup$ – PersonaA Oct 5 '15 at 23:19
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$\begingroup$ @PersonaA That's good that you found the statement unhelpful -- Empty set is most definitely not an element of $A$. $\endgroup$ – pjs36 Oct 5 '15 at 23:26
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$\begingroup$ As a side comment: $B \in A$, but $B \not\subseteq A$. $\endgroup$ – Brian Tung Oct 5 '15 at 23:26
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The solution is correct: it is not. In order for $B$ to be a subset of $A$, every element of $B$ must be an element of $A$. What are the elements of $B$? There is only one, $\varnothing$. What are the elements of $A$? $1$ and $\{\varnothing\}$. Neither of these is $\varnothing$, so $\varnothing\notin A$, and therefore $B\nsubseteq A$. Note that $\varnothing$, the element of $B$, and $\{\varnothing\}$, one of the elements of $A$, are not the same thing: $\varnothing$ is a set with no members, and $\{\varnothing\}$ is a set with one member, that one member being the set $\varnothing$.
answered Oct 5 '15 at 23:20
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$\begingroup$ But B is the set contain the null set right? $\endgroup$ – PersonaA Oct 5 '15 at 23:21
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$\begingroup$ @PersonaA: $B$ is the set whose only element is the null set. But the null set is not an element of $A$, so $B$ is not a subset of $A$. (You may be getting confused by the fact that $\varnothing$ is a subset of every set; it’s not an element of every set, however.) $\endgroup$ – Brian M. Scott Oct 5 '15 at 23:23
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$\begingroup$ Set theorists make everything a set. So what is are natural numbers in set theory? The usual definition is $0=\phi$ and $n+1=n\cup \{n\}$. $\endgroup$ – DanielWainfleet Oct 6 '15 at 0:56
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$\begingroup$ @user254665: Yes, and the set of natural numbers is then $\omega$, whose existence is guaranteed by the axiom of infinity. If one is using that formalism, $A=\{1\}$, and $B=\{0\}=1$. $\endgroup$ – Brian M. Scott Oct 6 '15 at 1:10
