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Can there be a set $A$ and a set $B$ such that $A\subseteq B$ and $A\ne B$ ?

While trying to find a solution to this question, I've found this answer which states:

An improper subset (usually denoted as $A\subseteq B$) is such that $A=B$ is allowed (but not mandated)

If that's true, can you give an example for such a case where $A\ne B$ ?

John
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2 Answers2

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Of course, $A= \{1\}$, $B =\{1,2\}$, will do. Or $A = \emptyset$ and $B$ any non-empty set, like $\{\emptyset\}$.

Henno Brandsma
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  • Aren't those examples of proper sets, since $A$ does not contain $2$? – John Jun 28 '20 at 11:48
  • @Noam there is no such thing as a proper set here. Both are sets and $A$ is a proper subset of $B$. – Henno Brandsma Jun 28 '20 at 11:50
  • Sorry, I meant a proper subset, Wikipedia defines a subset as proper if "there exists at least one element of B which is not an element of A" – John Jun 28 '20 at 11:53
  • @Noam $A\subseteq B$ means only that all elements of $A$ are in $B$ as well. – Henno Brandsma Jun 28 '20 at 12:07
  • Oh, it finally clicked! $\subseteq$ and $\subset$ aren't opposites! – John Jun 28 '20 at 12:21
  • @Noam Indeed. $\subset$ is used by some authors as a synonym for $\subseteq$, by others as an alternative for $A\subsetneq B$, which can both mean "all elements of $A$ are in $B$, but not $A = B$". The negation of $A \subseteq B$ is "there is some $a \in A$ that is not in $B$", denoted by $A \not\subseteq B$, e.g. – Henno Brandsma Jun 28 '20 at 12:24
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If you can prove for every $x$ in A is also in B but inverse is not true, you will finish. This is your proving motivation. For example: let $A={(1,2,3)}$ and $B={(1,2,3,4)}$ two sets. We have for all $x$ in $A$, $x$ is also in B. But there exist a $y$ in $B$ (which is $4$) which is not in $A$. So we have finally:

$A⊆B$ and $A≠B$

Absurdist
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