We discuss the notion of subset. These are sets that are part of another set.

Let #\blue A# and #\green B# be sets.

We say set #\blue A# is a **subset** of #\green B# if every element of #\blue A# is also an element of #\green B#.

We denote this by #\blue A\subseteq \green B# or by #\green B \supseteq \blue A#.

**Examples**

Let #\blue A = \{1, 4, 6\}# and #\green B = \{1, 2, 3, 4, 5, 6\}#.

Since #1\in\green B#, #4\in\green B#, and #6\in\green B#, we have #\blue A \subseteq \green B#.

The set #\orange C# is **not a subset** of #\green B# if at least one element of #\orange C# is not an element of #\green B#*. *

We denote this by #\orange C \not \subseteq \green B#.

In addition, let #\orange C = \{2.5, \pi, 5\}#.

The element #2.5# of #C# does not belong to #\green B # (and similarly for #\pi#). Therefore, #\orange C \not \subseteq \green B#.

If #\blue A# is a subset of the set #\green B#, then we also say that #\blue A# is **included** in #\green B#, that #\green B# **includes** #\blue A#, that #\blue A# is contained in #\green B#, or that #\green B# contains #\blue A#. The symbol #\subseteq# is also called **inclusion**.

Subsets can also be displayed in a Venn diagram as an oval within an oval.

Let #\blue A# and #\green B# be sets.

If #\blue A \subseteq \green B#, then this inclusion is visualized by drawing the oval corresponding to #\blue A# inside the oval corresponding to #\green B#.

If #\orange C \not\subseteq \green B#, then this non-inclusion is visualized by making sure the oval corresponding to #\orange C# does not lie in the oval corresponding to #\green B#. The elements in #\orange C# but not in #\green B# are drawn as points inside the oval of #\orange C# and outside the oval of #\green B#.

**Example**

As before, we take #\blue A = \{1, 4, 6\}#, #\green B = \{1, 2, 3, 4, 5, 6\}#, and #\orange C = \{2.5, \pi, 5\}#. In the Venn diagram, we see that #\blue A \subseteq \green B# because the oval of #\blue A# lies in the oval of #\green B#. Also, the points of #2.5# and #\pi# of #\orange C# do not lie in #\blue A# and are drawn outside the oval of #\blue A#.

Every set is a subset of itself because every element of a set #\blue A# is of course an element of #\blue A# itself.

If a set #\blue A# is a subset of a set #\green B# with #\blue A \neq \green B#, then we call #\blue A# a **proper subset** of #\green B#. It is written as #\blue A \subset \green B#. The symbol #\subset# is referred to as a **proper inclusion **or a **strict inclusion**.

**Example**

Consider sets #\blue A = \{1, 4, 6\}# and #\green B = \{1, 2, 3, 4, 5, 6\}#. We see that #\blue A \subset \green B#.

Recall that two sets #\blue A# and #\green B# are equal if and only if every element of #\blue A# is an element of #\green B# and conversely every element of #\green B# is an element of #\blue A#. Therefore, #\blue A=\green B# if and only if #\blue A \subseteq \green B# and #\green B\subseteq \blue A#.

**Example**

Let #\blue A# be the set #\{1, 2, 3\}# and #\green B# be the set #\{3, 2, 1\}#. Here #\blue A# is equal to #\green B#, because #\blue A \subseteq \green B# and #\green B \subseteq \blue A#.

Among the special sets of numbers, we have the inclusions #{\mathbb N}\subseteq {\mathbb Z}#, #{\mathbb Z}\subseteq {\mathbb Q}#, and #{\mathbb Q}\subseteq {\mathbb R}#.

The empty set is a subset of every set.

In order to prove this statement, we let #\blue A# be an arbitrary set. We will show that #\emptyset\subseteq A#. We do this by establishing that every element of #\emptyset# is also an element of #\blue A#. In the language of logic this means that the statement \[a \in \emptyset \Rightarrow a \in \blue A\] is always true. But this is trivially satisfied since #a \in \emptyset# is always false.

Containment of set is transitive in the sense that, for all sets #\blue A#, #\green B# and #\orange C#, we have \[(\blue A \subseteq \green B\text{ and }\green B \subseteq \orange C)\ \Rightarrow\ \blue A \subseteq \orange C\]

For example, if #\blue A = \{1, 2, 3\}#, #\green B = \{1,2,3, 4\}#, and #\orange C = \{1, 2, 3, 4, 6, 8\}#,

then \(\blue A \subseteq \green B\) and \(\green B \subseteq \orange C\), so #\blue A \subseteq \orange C#.

Consider the sets #A = \{ 3, 6, 20\} # and #B = \{3, 4, 6, 10, 13, 20\}#, is it true that #A \subseteq B#?

Yes

It is true that #A \subseteq B#. Because #A = \{ 3, 6, 20\} # is contained in #B = \{3, 4, 6, 10, 13, 20\}#. So every element of #A# is an element of #B#.