Can infinite sets differ finitely?

[Loren Booda]

Can two infinite sets differ by a finite number of elements?

 

[Moo Of Doom]

Take for example $A = \mathbb{N}$ and $B = \mathbb{N} \setminus \left\{ 1 \right\}$. Then $A$ and $B$ are both infinite sets, and they share all elements except for 1. Or is this not what you mean by "differ"?

 

[CRGreathouse]

Can two infinite sets differ by a finite number of elements?

I take your question as:

"Given two infinite sets A and B, is the symmetric difference $(A\cup B)\setminus(A\cap B)$ ever nonempty?"

in which case the above post gives an example of when this can occur.

 

[Loren Booda]

Please forgive my ignorance, but what does \ mean in this context?

 

[Dragonfall]

Let' say A = {0, 1, 2, ...} and B = {1, 2, 3, ...}. A and B are both infinite, but A and B "differ" by one element. Namely 0.

 

[CRGreathouse]

Please forgive my ignorance, but what does \ mean in this context?

Set subtraction (thus the LaTeX command \setminus). {1, 2, 3} \ {2} = {1, 3}.

 

[HallsofIvy]

I take your question as:

"Given two infinite sets A and B, is the symmetric difference $(A\cup B)\setminus(A\cap B)$ ever nonempty?"

in which case the above post gives an example of when this can occur.

Please forgive my ignorance, but what does \ mean in this context?

It's the "set difference". A\B is "All values that are in A but not in B". Think "A take away any members of A intersect B".

 

[Loren Booda]

Thanks, all.

 

[Loren Booda]

There isn't an inverse function to \, is there?

 

[Dragonfall]

Yes there is, the union. A \ B U B = A.

 

[secretman]

Not quite; only if B is contained in A.

 

[HallsofIvy]

Not quite; only if B is contained in A.

Only if B is a subset of A.

Using "is contained in" to mean "is a subset of" can cause an awful lot of confusion.