Set theory: Prove that [tex]\mathbb{N} \ne (\mathbb{N} \cup[/tex]...

[ksdhart2]

Hi all. I'm stumped on a problem again and I'm hoping someone here can help me out. The question says:

Prove that \(\displaystyle \mathbb{N} \ne (\mathbb{N} \cup \left\{\mathbb{N}\right\})\)

Here's what I've done so far, but I'm really not sure where to go next, or even if what I've done so far is any good...

• \(\displaystyle \vdash \forall I\left\{\left(I\in \mathbb{N}\right) \implies \left[\left(I+1\right) \iff \left(I\cup \left\{I\right\}\right)\right]\right\}\)
• \(\displaystyle \vdash \forall I \left\{ (I \in \mathbb{N}) \implies (I \ne I+1)\right\}\)
• \(\displaystyle \vdash (\mathbb{N} \in \mathbb{N}) \implies (\mathbb{N} \ne \mathbb{N}+1)\)
• \(\displaystyle \vdash \mathbb{N} \notin \mathbb{N}\)

Basically, I started by writing down the definition of I + 1 used by our book. Then I noted that there's no natural number equal to its successor. From that, I made a substitution to show that if the entire set of natural numbers was part of itself, then it would follow that it's not its successor. However, in a previous homework exercise, I proved that the set of natural numbers is not a member of itself.

So, yeah, like I said earlier, I don't know where to go from here. Any help would be greatly appreciated.

 

[stapel]

Hi all. I'm stumped on a problem again and I'm hoping someone here can help me out. The question says:

Prove that \(\displaystyle \mathbb{N}\, \ne \,(\mathbb{N} \,\cup\, \left\{\mathbb{N}\right\})\)

Here's what I've done so far, but I'm really not sure where to go next, or even if what I've done so far is any good...

• \(\displaystyle \vdash\, \forall\, I\,\left\{\left(I\,\in\, \mathbb{N}\right) \,\implies\, \left[\left(I\,+\,1\right)\, \iff\, \left(I\,\cup\, \left\{I\right\}\right)\right]\right\}\)
• \(\displaystyle \vdash\, \forall\, I\, \left\{ (I\, \in \,\mathbb{N}) \,\implies\, (I\, \ne\, I+1)\right\}\)
• \(\displaystyle \vdash\, (\mathbb{N}\, \in \,\mathbb{N})\, \implies\, (\mathbb{N}\, \ne\, \mathbb{N}+1)\)
• \(\displaystyle \vdash\, \mathbb{N}\, \notin\, \mathbb{N}\)

Basically, I started by writing down the definition of I + 1 used by our book....

...in a previous homework exercise, I proved that the set of natural numbers is not a member of itself.

To prove that two sets aren't equal, all you have to do is find one element from one of the sets which is not in the other set. You've already proved that the set of natural numbers, {N}, is not an element of N itself. Thus, you have proven that one element of the right-hand side of the original in-equation is definitely not in the left-hand side. So the sets cannot be equal.

 

[ksdhart2]

To prove that two sets aren't equal, all you have to do is find one element from one of the sets which is not in the other set. You've already proved that the set of natural numbers, {N}, is not an element of N itself. Thus, you have proven that one element of the right-hand side of the original in-equation is definitely not in the left-hand side. So the sets cannot be equal.

Wait... it's really that simple? Man. I was really overthinking this one. Revisiting the definition we used of the equals operator, I see now that what you're saying must be the case. Thanks a bunch!

 

[lookagain]

You've already proved that the set of natural numbers, {N}, is not an element of N itself..

N is already the set of natural numbers, so N with the braces around it is the set
containing the set of natural numbers as an element.