Logically Prove (A \cup B) \subseteq C

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In summary: Hi Ajsingh!I think your progression is correct except for premise 2. You say that x is an element of A, AND, x is an element of B. That can be true, but it doesn't have to be true. I recommend switching the AND to an OR. After you make that change, I don't see any fault in your logic. I also think it flows better if you switch premises 3 and 4. I know it's semantics but the progression seems better to me.
  • #1
ajsingh
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Homework Statement



To prove [tex]A \subseteq C and B \subseteq C[/tex] implies [tex](A \cup B)\subseteq C[/tex]

2. The attempt at a solution

I just wanted to know if my reasoning seems logical. Here is my attempt:

Assume [tex]A \subseteq C and B \subseteq C ... (1) [/tex]
Assume [tex]\forall x [ x\in A] and \forall x [x \in B] ... (2) [/tex]
Hence, from definition of [tex]\bigcup \forall x [x \in A \cup B ] ... (3)[/tex]

From (1) and defination of [tex]\subseteq, \forall x [ x \in A \Rightarrow x \in C and x \in B \Rightarrow x \in C ] ..... (4)[/tex]

Hence, [tex]\forall x [x \in A \cup B \Rightarrow x \in C] ..... (5)[/tex]

[tex]\Rightarrow (A \cup B) \subseteq C

[/tex]

Does that seem to flow logically?

Thanks
 
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  • #2
Hi Ajsingh!

I think your progression is correct except for premise 2. You say that x is an element of A, AND, x is an element of B. That can be true, but it doesn't have to be true. I recommend switching the AND to an OR. After you make that change, I don't see any fault in your logic. I also think it flows better if you switch premises 3 and 4. I know it's semantics but the progression seems better to me.

I also want to point out that unless specified in the problem, it might be enough to simply demonstrate that this proof is true. What if you assigned C={1,2,3,4,5}, A={1,2,3}, B={3,4}

A is a subset of C. B is a subset of C, A U B = {1,2,3,4} which also is a subset of C. Proved. I mention this, because it also shows a contradiction to your premise #2. Namely, 4 is an element of B but NOT an element of A.

Hope this helps!

Steve
 
  • #3
ajsingh said:

Homework Statement



To prove [tex]A \subseteq C and B \subseteq C[/tex] implies [tex](A \cup B)\subseteq C[/tex]

2. The attempt at a solution
You want to prove one set is a subset of another. The standard way of doing that is to say "if x in the first set" and show "therefore x is in the second set".

I just wanted to know if my reasoning seems logical. Here is my attempt:

Assume [tex]A \subseteq C and B \subseteq C ... (1) [/tex]
While this isn't, strictly speaking, wrong, you don't need to "assume" that- it is given.

Assume [tex]\forall x [ x\in A] and \forall x [x \in B] ... (2) [/tex]
Now you don't want to say that! For one thing, A and B might be disjoint- there might be no such x! What you want to say is "Assume [itex] x\in A\cup B[/itex].

Hence, from definition of [tex]\bigcup \forall x [x \in A \cup B ] ... (3)[/tex]
x doesn't have to be in both A and B in order to be in [itex]A\cup B[/itex]!

[quote\From (1) and defination of [tex]\subseteq, \forall x [ x \in A \Rightarrow x \in C and x \in B \Rightarrow x \in C ] ..... (4)[/tex]
You are going the wrong way. Starting from the assumption that [itex]a\in A \cup B[/itex] it follows that either [itex]x\in A[/itex] or [itex]x\in B[/itex].

Hence, [tex]\forall x [x \in A \cup B \Rightarrow x \in C] ..... (5)[/tex]

[tex]\Rightarrow (A \cup B) \subseteq C

[/tex]

Does that seem to flow logically?

Thanks
 

FAQ: Logically Prove (A \cup B) \subseteq C

1. What is the meaning of (A \cup B) \subseteq C?

The notation (A \cup B) \subseteq C means that the set A union B is a subset of the set C. This means that every element in A or B is also an element in C.

2. How can (A \cup B) \subseteq C be logically proven?

To logically prove (A \cup B) \subseteq C, we need to show that every element in A or B is also an element in C. This can be done by assuming an arbitrary element x in A or B, and then showing that x is also in C. This process is known as a direct proof.

3. Can (A \cup B) \subseteq C be proven using a contrapositive proof?

Yes, (A \cup B) \subseteq C can also be proven using a contrapositive proof. This involves assuming an element x that is not in C, and then showing that x cannot be in A or B. This proves that the statement "if x is in A or B, then it is also in C" is true.

4. How does De Morgan's law apply to (A \cup B) \subseteq C?

De Morgan's law states that the complement of a union is equal to the intersection of the complements. In this case, it means that the complement of (A \cup B) is equal to the intersection of the complements of A and B. This can be written as (A \cup B)^c = A^c \cap B^c. This can be helpful when proving (A \cup B) \subseteq C, as it allows us to work with the complements of the sets instead.

5. Are there other methods for proving (A \cup B) \subseteq C?

Yes, there are other methods for proving (A \cup B) \subseteq C, such as using a proof by contradiction or a proof by cases. In a proof by contradiction, we assume that (A \cup B) is not a subset of C, and then show that this leads to a contradiction. In a proof by cases, we consider different cases for the elements in A or B and show that in each case, the element is also in C. Ultimately, the method used will depend on the specific problem and the preferences of the mathematician.

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