I am trying to make sense of these. To me a is false because the set isn't empty. Is that correct?

b is true because the empty set is an element of that set.

c is false because the set the empty set isn't an element, it is a subset or proper subset.

d is true because the set the empty set is a member of that set.

e the set the empty set is false because it is not a proper subset I think

f false for the same reason

g false for the same

a) ∅ ∈ {∅}

b) ∅ ∈ {∅, {∅}}

c) {∅} ∈ {∅}

d) {∅} ∈ {{∅}}

e) {∅} ⊂ {∅, {∅}}

f ) {{∅}} ⊂ {∅, {∅}}

g) {{∅}} ⊂ {{∅}, {∅}}

I know not all these are correct, why is my logic wrong?

  • $\begingroup$ No, (a) is true: no matter what $a$ is, $a\in\{a\}$, because by definition $\{a\}$ is the set whose only element is $a$. $\endgroup$ – Brian M. Scott Oct 16 '13 at 0:16
  • $\begingroup$ e) is also true, since $A\subset B$ means that every element of $A$ is also an element of $B$. $\endgroup$ – Patterns_43 Oct 16 '13 at 0:18
  • 2
    $\begingroup$ (f) is true: the only element of $\{\{\varnothing\}\}$ is $\{\varnothing\}$, which is also an element of the righthand side. Whether (g) is true or false depends on whether you’re using $\subset$ to mean subset or proper subset; in the former case it’s true, and in the latter it’s false, since the righthand side is equal to the lefthand side. $\endgroup$ – Brian M. Scott Oct 16 '13 at 0:22
  • $\begingroup$ I don't even know how to parse "the set the empty set is false". What do you mean by "the set the empty set"? And how can a "set" be true or false? $\endgroup$ – bof Oct 16 '13 at 0:31
  • $\begingroup$ @bof {∅} that is the set the empty set, like {1} is the set containing 1, or {{1}} is the set the set containing 1. Anyways I meant to say the statement is false. $\endgroup$ – Paul the Pirate Oct 16 '13 at 1:19

Don't argue too much, do it mechanically instead. If you wipe out the enclosing $\{$ and $\}$ you will see a list of the elements of the set, e.g., if $A=\{a,b,\{a\}\}$ then after removing the braces, this $$a,b,\{a\}$$ is the list of elements of $A$, so $$a\in A,\quad b\in A\quad \{a\}\in A.$$

You'll get any subset of $A$ by first removing the braces to get the element-list, then choose some elements and put braces around your chosen elements, for example $$\{a,b\}\subset A,\quad \{a,\{a\}\}\subset A,\quad \{\{a\}\}\subset A.$$ So only (c) is wrong. For example, (f) is true: get the element list of $M=\{\emptyset,\{\emptyset\}\}$ by removing braces: $$\emptyset, \{\emptyset\},$$ chose the element $\emptyset$, (thus $\emptyset\in M$), put braces around your chosen element to get $\{\emptyset\}$, to yield a subset of $M$, that is $\{\emptyset\}\subset M$.

  • $\begingroup$ There is not an iota mention of the phrase 'proper-subset' in your answer. $\endgroup$ – Ubi hatt Jul 11 '18 at 18:46
  • $\begingroup$ Not only c, g is wrong too. A set can't be proper set of itself. $\endgroup$ – Ubi hatt Jul 12 '18 at 0:00

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