Recipe

Check whether A is a subset of B

Use when: you're given two sets $A$ and $B$ and must decide whether $A \subseteq B$ , and if so, whether it's proper

Goal: a settled verdict on $A \subseteq B$ , plus whether it's proper, backed by a full element check or one counterexample

Taught in Subsets and proper subsets

Quick recipe

Check every element of A against B
Decide the verdict from what you found
Check for a proper subset, once A is a subset

1
Take each element of $A$ in turn, and confirm it also belongs to $B$ . Testing every element is required, since a subset claim is itself a universal claim about the elements of $A$ .

Example

Is $A=\{3,6,9\}$ a subset of $B=\{1,2,3,6,9,12\}$ ? Check each: $3\in B$ , $6\in B$ , $9\in B$ .

Why: $A\subseteq B$ means literally every element of $A$ is also in $B$ , by definition, so nothing less than a full check settles it.
2
If every element of $A$ passed, $A\subseteq B$ . If even one element of $A$ is missing from $B$ , that element is a counterexample, and $A\subseteq B$ is false.

Example

All three elements of $A=\{3,6,9\}$ passed, so $A\subseteq B$ .
- every element of $A$ is in $B$ — $A\subseteq B$ holds.
- at least one element of $A$ is missing from $B$ — $A\not\subseteq B$ , witnessed by that element.
Why: one element of $A$ outside $B$ is already a complete disproof, the same way one counterexample disproves any universal claim.
3
Once $A\subseteq B$ is confirmed, check whether $B$ has an element $A$ lacks. If so, $A$ is a proper subset of $B$ ; if $A$ and $B$ have exactly the same elements, it is a subset but not proper.

Example

$B=\{1,2,3,6,9,12\}$ has $1$ , $2$ , and $12$ , none of which are in $A$ , so $A\subsetneq B$ .
- $B$ has an element not in $A$ — $A\subsetneq B$ .
- $A$ and $B$ have exactly the same elements — $A\subseteq B$ but $A\not\subsetneq B$ .
Why: "proper" adds the extra requirement that $A$ and $B$ actually differ, on top of every element of $A$ landing in $B$ .

Common mistakes

Stopping after checking only some elements of $A$ and declaring $A\subseteq B$ without finishing the check.
Checking elements of $B$ against $A$ instead of elements of $A$ against $B$ -- the direction matters.
Concluding $A\subsetneq B$ just because $A$ and $B$ look different, without confirming $A\subseteq B$ first.
Treating a shared element between $A$ and $B$ as evidence, when only an element of $A$ missing from $B$ actually disproves the claim.