This is from Discrete Mathematics and its Applications enter image description here

What is the symbol used in 9c, 9d, 9f, 10c, 10f, 10g? I looked through the chapter section and the closest symbol I saw to this is the subset, which is

DEFINITION 3. The set $A$ is a subset of $B$ if and only if every element of $A$ is also an element of $B$. We use the notation $A \subseteq B$ to indicate that $A$ is a subset of the set $B$.

We see that $A \subseteq B$ if and only if the quantification $$\forall x(x \in A \to x \in B)$$ is true. Note that to show that $A$ is not a subset of $B$ we need only find one element $x \in A$ with $x \not\in B$. Such an $x$ is a counterexample to the claim that $x \in A$ implies $x \in B$.

Is it just a typo for subset? Thats what I originally thought. However via my use of an implication(trying to apply what I learn :) ), I came up with if the symbol is a typo, it will be used in a single place or the supposed actual symbol, the subset, will not be used in the surrounding proximity(page). If I assumed the hypothesis to be true, then my conclusion and my implication is false because the subset does appear in the near proximity(9g) and this symbol is used in multiple locations(all the ones I described). Therefore the hypothesis is false(reached a contradiction), and the symbol is not a typo. Is that correct logic?

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    $\begingroup$ It should be noted that some authors use $\subset$ for $\subseteq$ and denote $\subsetneq$ for a proper subset. $\endgroup$
    – Hugh
    Jan 14, 2015 at 7:58

2 Answers 2


The textbook is not making typos. The symbol $\subset$ is called a proper subset. For example, $$[0,1] \subset [0,2],$$ but $$[0,2] \not\subset [0,2].$$

However, the symbol $\subseteq$ means that a set can be contained in itself as well, for example, $$[0,1] \subseteq [0,2]$$ and $$[0,2] \subseteq [0,2].$$

This page describes a little more on the difference between subset and proper subset.

  • $\begingroup$ For this [0,1]⊆[0,2] meaning [0,1] is a subset of [0,2], doesn't [0,2] have to have every element in [0,1]? Which it doesn't because of the 1 in [0,1]? $\endgroup$ Jan 14, 2015 at 6:49
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    $\begingroup$ I am describing a continuous interval $[0,2]$, which does indeed contain the subinterval $[0,1]$; hence, $[0,1] \subseteq [0,2]$. (Note that an interval is a type of set on a real line.) I am not describing a collection set of elements $\{0,2\}$, which does not contain the $1$ as you said; hence, $\{0,1\} \not\subseteq \{0,2\}$. $\endgroup$
    – Cookie
    Jan 14, 2015 at 6:56
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    $\begingroup$ oh thanks that makes sense. By the way, how did you embed that subset symbol into your comment? Every time, I have to copy and paste.... $\endgroup$ Jan 14, 2015 at 7:07
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    $\begingroup$ I used the MathJax code of \subset for "$\subset$" and \subseteq for "$\subseteq$". Please see meta.math.stackexchange.com/questions/5020/… for how to use the MathJax code needed to type math symbols on this site. $\endgroup$
    – Cookie
    Jan 14, 2015 at 7:14

The symbol $\subset$ denotes proper subset.

$$A\subset B \iff \forall x (x \in A \to x\in B)\wedge \exists x (x\in B \wedge x\not\in A)$$

Which means that all the elements in the proper subset are elements of the set, and there are elements in the set which are not in the proper subset.   A set cannot be a proper subset of itself.

Where as the symbol $\subseteq$ denotes the subset or equivalent, commonly referred to as just subset.   A set can be its own subset.

$$A\subseteq B \iff \forall x (x\in A\to x\in B)$$

Likewise the symbol $\supset$ and $\supseteq$ denote proper superset and superset.

$$A\supseteq B \iff \forall x (x\in A\leftarrow x\in B)$$

$$A\supset B \iff \forall x (x \in A \leftarrow x\in B)\wedge \exists x (x\in A \wedge x\not\in B)$$

And of course we have set equivalence, $\equiv$

$$A\equiv B \iff \forall x (x\in A \leftrightarrow x\in B) $$

These symbols are analogous to the order symbols $<$, $\leq$, $\geq$, $>$, and $=$

  • $\begingroup$ So proper subset is basically subset but it can't be the same set? $\endgroup$ Jan 27, 2015 at 6:11
  • $\begingroup$ Yes, just so, @committedandroider $\endgroup$ Jan 27, 2015 at 11:34

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