Basic set theory with quantors question

[Smalde]

This is no homework for me.
I am working as a teaching assistant in a lecture about logic and discrete structures for Informatics students. This should be a piece of cake, but I am not exactly sure of the logic behind.

1. Homework Statement

Translate into words
∃c . ∀a ∈ A . ∀b ∈ B . ¬(a = b) ∨ c = b

The Attempt at a Solution

So obviously this means that there is a c such that for all a element of A and all b element of B either a is different than b or c is b or both.
The problem is that they also said that this means that the intersection of A and B has maximally one Element, 1 ≥ |A ∩ B|.My questions is how exactly can one conclude the last part. ∃c means that there is at least one, not necessarily only one.

Obviously for the statement to be true whenever a equals b c has to equal b and thus a, so either ¬(a = b) or a = b = c. But if there can be two cs this statement can be true for two different bs and as and so the intersection needs not be maximally 1...

Could you help me out?

PS: I am not sure if this has to be here... I can move it to another sub-forum if necessary.

 

[Stephen Tashi]

But if there can be two cs

Suppose $A = \{1,2,3\}$, $B = \{4,2,3\}$. Can you find any c that satisfies the statement in the problem?

 

[Smalde]

Suppose $A = \{1,2,3\}$, $B = \{4,2,3\}$. Can you find any c that satisfies the statement in the problem?

Well 2 and 3. That's what I do not understand. I was told that the "there exists" symbol means that there exists at least one of something.

 

[Smalde]

Well 2 and 3. That's what I do not understand. I was told that the "there exists" symbol means that there exists at least one of something.

Ok, wait.

I think I get it now: there can exist more than one, but all ought to fulfill the statement in every one case. Is it so?

 

[Stephen Tashi]

Well 2 and 3.

Neither 2 nor 3 works.

For example c = 2 does not work for the case a = 3, b = 3.

 

[Smalde]

Ok, wait.

I think I get it now: there can exist more than one, but all ought to fulfill the statement in every one case. Is it so?

Neither 2 nor 3 works.

For example c = 2 does not work for the case a = 3, b = 3.

Yes, that's what took me so long to understand. Those cs have to work for all possible cases where a equals b. In other words, in this case, there can only be one such c.

THANKS!