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Does every subset start with an empty set?

If $A = \{a,b\}$, then the subsets of $A$ are $\{a\},\{b\},\{a,b\}$.

Not sure if that is correct.

Daniel W. Farlow
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user4744409
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    Do you mean to say "the empty set should always be listed as a subset" when you say "start with an empty set"? If so, the answer is yes. – Cameron Williams Apr 03 '15 at 03:39
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    there's no necessary "sequence" nature to subsets. Your question is ambiguous at best. – Memming Apr 03 '15 at 03:42
  • @CameronWilliams I agree with your correction concerning the statement of the question. I don't think OP meant to imply a sequence/ordering on the subsets of $A$, but OP has not responded to yours or Memming's comments--so who knows? Either way, these comments and mine and Michael's answers will hopefully resolve OP's question. – Daniel W. Farlow Apr 03 '15 at 03:53
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    I'm voting to close this question as a duplicate of is the empty set a subset of every set? because this question, in its essence, is a duplicate of that question. – Daniel W. Farlow Apr 03 '15 at 04:04

3 Answers3

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The empty set $\emptyset$ is a subset of every set, including the set $A$. The subsets of $A$ are $\emptyset$, $\{a\}$, $\{b\}$, and $\{a, b\}$.

Michael Albanese
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From the definition of subset containment:

If, for every $a \in A$, we have $a \in B$, then we say $A \subseteq B$.

To test whether $\emptyset \subseteq S$:

For each $x \in \emptyset$, we need to make sure $x \in S$. But $x \in \emptyset$ is always false, so the implication is vacuously true; thus, $\emptyset \subseteq S$, for all sets $S$ (even if $S = \emptyset$).

pjs36
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  • I think the OP literally has everything now necessary to answer his\her question. Variety of answers here is nice. – Daniel W. Farlow Apr 03 '15 at 03:58
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    Indeed! Resorting to the definition is a little dodgy. If I had to guess, the issue is either 1) Not using the definition, or 2) Not being comfortable with vacuously true statements. – pjs36 Apr 03 '15 at 04:02
  • Perhaps. I voted to close because I think the real question is whether or not an empty set is a subset of every set. But the OP refuses to clarify for some reason. :/ – Daniel W. Farlow Apr 03 '15 at 04:05
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Adding to Michael's answer: are you familiar with what a power set is (briefly, it is the set of all subsets of a set)? If so, then you would know that $|\mathcal{P}(A)|=2^n$, where $n$ denotes the number of elements in $A$.

For your problem, you have $A = \{a,b\}$. Hence, $n=2$ and so you know $|\mathcal{P}(A)|=2^2=4$. Thus, there are four subsets of $A$. You listed three of them: $\{a\},\{b\},\{a,b\}$. The one you forgot was, indeed, the empty set.

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Daniel W. Farlow
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