I am a bit confused about the concept of proper subsets,precisely whether to include one or both of the void set and the set itself.

An extract from my module goes like this :

Obviously,every set is the subset of itself and the void set $\emptyset$ is the subset of every set. These two subsets are called improper subsets.

It also includes a theorem which states that "Let A be a finite set having n elements. Then the total number of subsets of A is ($2^n$) and the number of proper subsets of A is ($2^{n}-1).$"

Then again in a sample solution of this problem "If A = {a,b,c},then the number of proper subsets of A is ?"

Total no of subsets of {a,b,c} = $2^3$ = 8. But each set have two improper subset, so number of improper subsets are 6.

Is this solution correct ? If so please explain the concept.

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    $\begingroup$ please give reference of the definition and theorem in what book it appeared? thanks. $\endgroup$ – Hamza Rashid May 15 '18 at 19:58

Calling $\emptyset$ and $A$ "improper" subsets of a set $A$ is not universal, and it is confusing in this case, because the meaning of "proper" is not the same. It is standard to say that $S$ is a proper subset of $A$ if (and only if) every element of $S$ is an element of $A$, but $S$ is not equal to $A$, i.e., at least one element of $A$ is not in $S$. Under this definition, $\emptyset$ is a proper subset of every nonempty set, even though it is "improper" according to the convention you were also given. Just remember that mathematical terminology varies and isn't always logical. Here "improper" does not mean "not proper". (For this reason I would personally not use the convention of calling the empty set "improper".)

When finding all proper subsets, you should count the empty set.

  • $\begingroup$ Hm,so it's the matter of definition,so if I ever encounter problem like the one I mentioned in my question,I should use the theorem and 7 should be the correct answer right ? $\endgroup$ – Quixotic Nov 10 '10 at 5:49
  • $\begingroup$ Yes, it would be 7, consistent with the theorem you were given. But really the best source of wisdom is your teacher, because to some extent he or she may decide on the terminology used in a given context. $\endgroup$ – Jonas Meyer Nov 10 '10 at 5:52
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    $\begingroup$ This reminds me of another case of seemingly contradictory terminology: $X$ is dense in $X$ does not mean $X$ is dense in itself. $\endgroup$ – Jonas Meyer Nov 10 '10 at 5:59
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    $\begingroup$ @Debanjan: It is probably worth repeating, this time more to the point: Check with your teacher to be sure, because given the context it seems he or she may not consider the empty set "proper" in the local terminology, even though that contradicts the theorem you were given. There is no absolute right and wrong in definitions, even though "proper" has its well established meaning 99% of the time. (Similarly, I once saw two professors have a mild disagreement over the meaning of "equivalent metrics".) $\endgroup$ – Jonas Meyer Nov 10 '10 at 6:19
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    $\begingroup$ @Jonas Meyer: do you remember what that disagreement was? I'm curious. $\endgroup$ – Carl Mummert Dec 22 '10 at 20:44

Given a set $A$, a proper subset is any set $B$ such that $B\subseteq A$ and $B\neq A$; that is, $B$ is contained in $A$ but is not equal to $A$. This is denoted by $B\subset A$ in some texts.

So while $A$ is a subset of itself, it is not a proper subset of itself. And this is true for any set, even the empty set (or void set, as you call it).

Speaking of which, the empty set $\emptyset$ is not only a subset of any set, but also a proper subset of any non-empty set.

Edited to include a solution to OP's new question:

You have the answer in front of you. If $A$ has cardinality $n$, then the number of subsets is $2^n$ and the number of proper subsets is $2^n-1$, because the only set we have to "throw out" is $A$ itself in order to get all the proper subsets.

So if $A=\{a,b,c\}$, then there are $2^3-1=7$ proper subsets. We can just list them:









A set A is a subset of a set B if every element of A is also an element of B.

A set A is a proper subset of a set B if A is a subset of B and there is at least one element of B that's not an element of A.

Thus, the void set is a subset of all sets, and it's a proper subset of every set except itself.

Also, notice that we notate the void set using $\emptyset$, not $\phi$.


The basic idea here is that of Vacuous Truth. The universal quantifier applied to the empty set is, by definition, true. In symbols, $\forall{x}{\in}\emptyset \, P(x)$ is true, regardless of the statement $P(x)$ (see also this Wikipedia page). This is mainly defined this way for convenience, because otherwise you would always have to consider the empty set as a special case every time you used the universal quantifier.

This apples to subsets because a set $A\subseteq B$ by definition means $\forall x\in A \, x\in B$. If $A=\emptyset$, then here $P(x)=x\in B$, and we see from the above that this must be true, i.e., $\emptyset \subseteq B$ for any set $B$.

By the way, "proper subset" almost always means $A\subset B;\,A\neq B$, so of course $\emptyset$ is a proper subset of any set, except for itself.

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    $\begingroup$ I disagree that it is "defined this way for convenience." One way to see that $\forall{x}{\in}\emptyset, P(x)$ must be true for all $P$ is to consider its negation, which would assert that there exists $x\in\emptyset$ such that...OK, we can stop there. $\endgroup$ – Jonas Meyer Dec 22 '10 at 20:50
  • $\begingroup$ @Jonas: I see what you mean. The Wikipedia article (and at least one of my professors) seem to imply that it isn't as clear-cut as that, though. $\endgroup$ – asmeurer Dec 22 '10 at 22:02

By definition, a set A is said to be a proper subset of another set B if and only if A is a subset of B and A is not equal to B. In other words, A is said to be a proper subset of B if B isn't a subset of A. Hence it follows that the empty set is a proper subset of every set.

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    $\begingroup$ Is it also a proper subset of itself? If the answer is yes, then we have that the empty set is not unique, and therefore should be named "an empty set" instead; if the answer is no then it is not a proper subset of every set. $\endgroup$ – Asaf Karagila May 6 '11 at 16:28

protected by Asaf Karagila Jun 3 '16 at 9:11

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