Can anyone help me clear this up.

  1. Is {0, 1} ∈ {0, {1}}
  2. Is {0, 1} ⊆ {0, {0, 1}}
  3. Is b ∈ {a, {a, b}}
  4. And finally {a,b} ⊂ {a,a,b}

So I understand that a subset and proper subset (set without line under it) are different, because in a proper subset, A cannot be equal to B if A is a proper subset for B.

I think I'm mostly getting confused with how the set's break down to determine their elements, like for example in problem 1 is the sets elements {0, empty set, {{1}}, {{empty set}}}? So the cardinality would be 2?

  • $\begingroup$ Basic difefrence: $\in$ and $\subseteq$ are different relations. For an element $a$ of a set $A$ we have $a \in A$. For a subset $B$ of a set $A$ we have $B \subseteq A$. $\endgroup$ – Mauro ALLEGRANZA Oct 4 '17 at 15:49
  • $\begingroup$ $\{ 0,1 \}$ is not and element of $\{ 0, \{ 1 \} \}$ because the first is not "listed" between the elements of the second. The elements of the second are: $0$ and $\{ 1 \}$. $\endgroup$ – Mauro ALLEGRANZA Oct 4 '17 at 15:51
  • $\begingroup$ only 4 is true and the rest are not $\endgroup$ – Guy Fsone Oct 4 '17 at 15:51
  • $\begingroup$ The condition for $B$ being a subset of $A$ is that every element of $B$ is also an element of $A$: for every $x$, if $x \in B$, then $x \in A$. $\endgroup$ – Mauro ALLEGRANZA Oct 4 '17 at 15:52
  • $\begingroup$ The elements of 1. are two: $0$ and $\{ 1 \}$. Thus $\{ 0,1 \}$ is not an element of the set. $\endgroup$ – Mauro ALLEGRANZA Oct 4 '17 at 15:53
  1. The elements of {0, {1}} are 0 are {1}. Neither of these are equal to {0,1}, so {0,1} is not an element of {0, {1}}.

  2. To determine whether {0,1} is a subset of {0, {0,1}}, you must ask whether every element of {0,1} is an element of {0, {0,1}}. This is not the case, since 1 is an element of {0,1}, but 1 is not an element of {0,{0,1}}; the elements of {0,{0,1}} are 0 and {0,1}, neither of which are 1.

  3. The elements of {a,{a,b}} are a and {a,b}. Are either of these equal to b?

  4. It is true that {a,b} is a subset of {a,a,b}. The elements of {a,b} are a and b, and both of these are in {a,a,b}. However, it is also true that {a,a,b} is a subset of {a,b}, for the same reason. The fact that a appears twice is irrelevant; it only matters whether or not it is there. To summarize, {a,b} is a subset, but not a strict subset, of {a,a,b}.

  • $\begingroup$ Thank you for your clear explanation $\endgroup$ – Mike Oct 4 '17 at 15:57
  • $\begingroup$ for question 3, had the question been {b} ∈ {a, {a, b}}, would this have been true? Because the elements of the second set are a and {a,b}, which {b} is a part of the element {a,b}. $\endgroup$ – Mike Oct 4 '17 at 16:26
  • $\begingroup$ No, because {b} is not equal to {a,b}. $\endgroup$ – Mike Earnest Oct 4 '17 at 16:30

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