# Does {a^(-2)a, a^(-1)} form a set?

1. Sep 24, 2008

### tgt

Does {a^(-2)a, a^(-1)} form a set?

2. Sep 24, 2008

### CRGreathouse

Re: sets?

I'm not sure what you mean. If a is just a member of some arbitrary group, then $a^{-2}a=a^{-1}$ and so $\{a^{-2}a,a^{-1}\}=\{a^{-1}\}.$ But it's still a set.

But perhaps you mean something else?

3. Sep 24, 2008

### tgt

Re: sets?

A set cannot contain repeated elements so if a is a member of a group then
[itex]\{a^{-2}a,a^{-1}\}[itex] wouldn't be a set?

4. Sep 24, 2008

### poutsos.A

Re: sets?

According to what axiom in axiomatic set theory you created that set????

5. Sep 25, 2008

### Tac-Tics

Re: sets?

In regular notation, we would say that {x} = {x, x} = {x, x, x}, etc.

It's not that a set can't contain duplicate entries... it's that you aren't allowed to ask "how many copies" of something are in a set.

Here's an example of something quite similar. The image (sometimes called the range) of a real function f, is the set

$$\{f(x) | x \in R\}$$

Let f be the sine function. The image is then $$\{sin(x) | x \in R\}$$.

Notice that this means that $$sin(0), sin(\pi), sin(2\pi), sin(3\pi)$$, etc, are all in the image. But they are all 0! Does that mean that the image isn't a set, since 0 is in there many times? Not a bit. But the extra entries are redundant in this particular case.

Another way to make this clearer is to think of the notation $$X = \{x_1, x_2, ...\}$$ as a shorthand for
$$x_1 \in X, x_2 \in X, ...$$