5

Prove or disprove: if A is a subset of B and B is not a subset of C, then A is not a subset of C.

I know it is false for the counter example:

A = {1, 2}

B = {1, 2, 3, 4}

C = {1, 2, 6, 5}

How can I prove that mathematically?

Arturo Magidin
  • 398,050
Franceia
  • 159
  • 3
    All you need to do to prove that something is false is to provide a counterexample. It seems you've done that. – KReiser Mar 20 '12 at 04:03

4 Answers4

7

You just did; it's called a Counterexample.

You simply note that these particular $A$, $B$, and $C$ satisfy the hypotheses ("$A$ is a subset of $B$ and $B$ is not a subset of $C$"), but fail to satisfy the proposed conclusion ("$A$ is not a subset of $C$"). So the proposed implication cannot always hold. This disproves the assertion.

Arturo Magidin
  • 398,050
0

There exist an element x that is a member of set A and since set A is a subset of set B, then that element x is also a member of set B.

Since set B is not a subset of set C that there does not exist an element x which is a member of set C, therefore element x is not in set C

Therefore element x is a member of set A but not an element in set C therefore set A is not a subset of set C.

Neville Forrester
  • 1
-1

A is the subset of B given: B is not the subset of C given: Let x be an arbitrary element such that x belongs to A than we have to show that x doesn't belong to C =>xEA (:. given) =>xEB (A is subset of B) =>x doesn't belong to C (B is not the subset of C) but x is an arbitraray element of A hence proved that A is not the subset of C

Azhar Ali
  • 1
  • 1
    This is incorrect. "$A$ is not a subset of $C$" does not mean that elements of $A$ are not elements of $C$. "$A$ is not a subset of $C$" means "there exists at least one $x$ such that $x\in A$ and $x\notin C$. Your implication that $x$ does not belong to $C$ because it belongs to $B$ is incorrect, based on that misunderstanding of what "not a subset" means. – Arturo Magidin May 16 '12 at 05:49
  • You correctly proved that "If $A$ is a subset of $B$, and $B$ is disjoint from $C$, then $A$ is disjoint from $C$." However, the problem is different. – Arturo Magidin May 17 '12 at 18:01
-1

You can say that if $A$ is included in $B \cap C$ then it is obviously a subset of $B$ and $C$ by definition of the intersection. But $B$ need not to be a subset of $C$ for $B \cap C$ to be non empty hence for $A$ to be non empty. Your example illustrate this quite well

RiezFrechetKolmogorov
  • 240