Definition of $-A$?

Question

My textbook (independently written by a professor at my university) states the following definition:

Let $A \subset \mathbb{R}$. We define the opposite of $A$ as the set of all the opposite elements: $-A = \{x \in \mathbb{R} : -x \in A\}$

The book than gives an example, if $A = [-1, 3)$ then $-A = (-3, 1]$. However the bit I can't quite understand is this: take for example the points $x = \frac{1}{2}, y = 2$, both are real numbers but only one satisfies the condition since $-x \in A, -y \notin A$.

Shouldn't the definition be $-A = \{-x \in \mathbb{R} : x \in A\}$ or is it just saying the same thing?

Answer 1

I don't see what confusion your example results in. Perhaps you've done a mistake. You have $-x\in A$ and $x\in -A$ because $-3<x = 1/2 \le 1$, also you have $-y \notin A$ and $y\notin -A$ because $1 < y = 2$.

Your alternate definition may look fine on the surface and it suffices if you understand it correctly and it's equivalent.

However there's a problem/gotcha in that kind of construct in general. Let suppose we used another transform $\phi$ then your style of definition would be $\phi A = \{\phi x: x \in A\}$, but that raises the question of given a $\phi x$ candidate for membership, which $x$ do we test for membership in $A$ to decide if $\phi x\in \phi A$? If it's a bijective operation like $-$ this is not a real problem.

Actually your professors notation has the same weakness in that it is wrong if you just write $\phi A = \{x: \phi x \in A\}$, this is not what is meant, it just happens to work for negation.

The general definition of $\phi A$ is instead something like $\phi A = \{u: \exists x\in A(u=\phi (x))\}$ - somewhat similar to your, it's the same if you apply the understanding that $\{\phi x: x\in A\}$ means that given a $\phi x$ there exists a solution $x\in A$.

Answer 2

Think for a moment of another problem:

$\textit{Solve, in the field of reals, this equation } \mathrm{ ax^2+bx+c=0} \textit{, where } \mathrm{ a, b, c} \in \mathbb{R} \textit{ for the unknown } \mathrm{x}$

If you want to avoid all such wording without loosing any information, you can write:

$$\{x \in \mathbb{R}: ax^2+bx+c=0\}\textrm{, where } a, b, c \in \mathbb{R}$$

You can see that in the previous symbol what happen to be before the colon (a pipe can also be used) says what among the symbols that follow are to be considered the unknowns (also free variables) and what is their domain. Other symbols are to be considered parameters whose meaning must be specified elsewhere (explicitly as is the case for $a, b, c$ or not when obvious as with $+, =, 0$). What follows the colon, in this case the equation, is to be considered a logical predicate: in this case it must be read:

$$\textit{find all } \mathrm{x}\textit{'s for which it is true that }\mathrm{ax^2+bx+c=0}$$

Now reverting to your problem the predicate is $-x\in A$ and the unknown is $x$ that must be chosen among the reals, so this:

$$\{x\in \mathbb{R}: -x \in A\}$$

says,

$$\textit{find all } \mathrm{x}\textit{'s among the reals for which it is true that} -x\in A$$

and this:

$$\mathrm{-A=}\{x\in \mathbb{R}: -x \in A\}$$

says,

$$\textit{... and call }\mathrm{ -A } \textit{ the set of such } \mathrm{x}\textit{'s}$$

So you cannot put anything more before the colon, other than the declaration of the free variables and their domains.

In the end you confusion is resolved when you do not stop after your consideration: $$\mathrm{2 }\in \mathbb{R}\textit{ and }\mathrm{ -2 }\notin\mathrm{ A}$$

but keep going with:

$$\textit{... and the predicate being not satisfied means that }\mathrm{ 2 }\textit{ does not belong to }\mathrm{ -A}$$