In this chapter, we deal with sets and perform various operations on them. We will also get acquainted with the notion of the “relation.” For the convenience of the reader, a number of basic properties of sets is first collected, which will be then used in the remainder of this chapter to help solve problems. The same applies to the subsequent chapters.
The union of two (or more) sets denoted by A ∪ B is the collection of elements belonging to at least one of the sets A and B.
The difference of two sets denoted by A ∖ B is the collection of elements belonging to A and not belonging to B. This operation is also called the relative complement.
The complement set A′ is the collection of all elements of a certain space X not belonging to A: A′ = X ∖ A. Obviously A ∪ A′ = X.
The subset of a given set A is a certain set B composed of the elements of A only (not necessarily all of them). It is denoted by B ⊂ A. The empty set, i.e., the set not containing any elements, is a subset of any set: ∅ ⊂ A. Naturally also A ⊂ A.
Reflexivity: for each a ∈ A there is $(a,a)\in \mathcal {R}$.
Naturally, such a figure is drawn for illustrative purposes only. It would be reasonable to perform other properly modified sketches for several various configurations of sets (e.g., when one or all of them are disjoint). However, even with the figure drawn as it is allows us to gain some intuition. These types of drawings are called Venn diagrams.
At the end, it is worth mentioning that this type of proof can be easily and in a relatively simple way performed with the use of basic logic laws. To this end, let us create a table containing all possible logical values for three sentences: x ∈ A, x ∈ B, and x ∈ C. (In the present case, there are 8 possibilities, but one can imagine that with more complicated expressions and with more sets, such a table will eventually grow.) To demonstrate (1.1.1), one has to include the following sentences too: x ∈ A ∖ C, x ∈ A ∖ B, x ∈ B ∖ C, and x ∈ A ∖ B ∪ B ∖ C. Their Boolean values are derived from the first three sentences and the definitions of operations, such as “∖” and “∪.” As usual the symbol “1” means “truth,” and the symbol “0” means “false.”
x ∈ A | x ∈ B | x ∈ C | x ∈ A ∖ C | x ∈ A ∖ B | x ∈ B ∖C | x ∈ A∖B∪B∖C |
---|---|---|---|---|---|---|
0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 1 |
1 | 0 | 1 | 0 | 1 | 0 | 1 |
1 | 1 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 0 |
The first equation to start with can be called the associativity of the symmetrical difference. The second one is just the distribution of the intersection over symmetrical difference.
We then have the expression, which is the sum of four sets. Two of them are underlined and will be referred to below. This is not yet the final result, but a more perceptive eye could consider the proof as practically complete. Why? Well, it is because the right-hand side is fully symmetric with respect to the exchange of the set names A, B, and C. Furthermore, the operation Δ is symmetric too. Therefore, on the left-hand side, instead of A Δ(B ΔC) one could equally write (B ΔC) ΔA. Then, using the symmetry, one would easily justify that this expression must be equal to (A ΔB) ΔC and thus equality (1.1.16) holds.
In examples of that kind, the main difficulty lies in selecting transformations that one needs to perform on the expressions. In general, they can be done in many ways. The advice is such that one should always keep in mind the objective. Therefore, the formulas ought not be “mechanically” transformed but only in such a way as to obtain the desired structures.
A solution of this kind of task usually consists of two steps. First, using the appropriate illustration and one’s imagination, we specify the preliminary thesis (i.e., in our case the concrete form of B and C). The second step is a strict demonstration of the formulated thesis.
All elements of the set {0}×]0, 1[ belong to C.
Any element not belonging to {0}×]0, 1[ does not belong to C either.
For x ≠ 0, the inequality y > tx ^{2} can be rewritten as t < y∕x ^{2}. Can it be satisfied for any t ≥ 1? Of course not because the right-hand side is fixed (it is a concrete number) and t on the left-hand side can be given an arbitrarily large value. This means that there exists such t ≥ 1, that (x, y) ∉ A _{t}. As a consequence (x, y) cannot belong to all sets, and hence, nor to their intersection C.
For y ≥ 1, one is not able to satisfy inequalities y < −tx ^{2} + 1 because the number on the right is at most equal to 1 (remember that t is positive), and the inequality is strict. Such a point inevitably lies beyond A _{t} for t ≥ 1, and, therefore, is not in C.
For y ≤ 0, our reasoning is carried out in a similar way. This time the inequality y > tx ^{2} cannot be satisfied, as the number on the right-hand side is nonnegative.
The conclusion is: no point lying outside the set {0}×]0, 1[ belongs to C, and any point belonging to it belongs also to C. In this way, the equality (1.2.6) has been shown.
The question arises as to whether this number is also a supremum or at most one of the many (i.e., infinitely many) upper bounds. One knows that a supremum is the smallest upper bound, if it exists. In the space of real numbers it is known that it certainly exists, but not necessarily in the space of rational numbers. As an example of such a situation, one can consider the set $\{q\in \mathbb {Q}\;\;|\;\; q^2<2\}$, the extremal bounds of which (equal to $\pm \sqrt {2}$) do not belong to the space $ \mathbb {Q}$ (which means, thatde facto neither the supremum nor infimum exists).
A certain puzzle for the reader may constitute the question, how we knew from the beginning that the expressions (1.3.2) and (1.3.5) should have been transformed in a way to extract 3∕5 and − 1. Well, this comes from the formula for x and from our intuition: one can see without any calculations that because of − 1 in the numerator and + 1 in the denominator the fraction is less than 3∕5 and approaches this value for very large y. So if the value 3∕5 is extracted, it remains only to determine whether the second term is positive or negative. The identification of the sign is generally much simpler than finding a specific value. The same applies to the infimum, where the substitution y = 0 is conspicuous, since |y| takes then the smallest value.
In mathematics, but also in physics, a particularly important role is played by the so-called equivalence relations. Their definition is going to be recalled below. A relation $\mathcal {R}$ is called the “equivalence relation” if, and only if, it satisfies the following three conditions:
For each x ∈ X one has $(x,x)\in \mathcal {R}$. The relation with this property is called reflexive.
holds. As discussed above, in this case one is talking about the symmetric relation.
Such relation is called transitive.
In conclusion, one finds that the relation defined in the text of the present problem is the equivalence relation. As the reader certainly knows from the lecture, the equivalence classes form the partition on $\mathbb {Z}$. These classes will be found below.
Reflexivity. Naturally each element $x\in \mathbb {R}$ is $\mathcal {R}$-related to itself, because the condition ϕ(x) = ϕ(x) is obviously satisfied.
Symmetry. From ϕ(x) = ϕ(y) it follows that ϕ(y) = ϕ(x), so the relation is symmetrical.
Then a set of equivalence classes has been found and what remains is to create a graph. It is very simple because one only has to mark on the plane all pairs (x, y), for which the coordinates satisfy the equation (1.4.18). This is, however, equivalent to the logical disjunction of (1.4.19) and (1.4.20). The graph will then consist of two straight lines and the circle, mentioned above. Again one recognizes here the elements on which the attention was drawn in the previous exercise: the graph contains the line y = x (reflexive relation), and is symmetric with respect to this line (symmetric relation).
The classes found earlier can be very easily read off from the drawing. All one has to do is to choose some x _{0} on the horizontal axis and draw from it an auxiliary (vertical) line: x = x _{0}. This line, depending on the value of x _{0}, crosses the graph in two, three, or four points. The coordinates y of these points are just elements that form the class $[x_0]_{\mathcal {R}}$. Naturally among them we will find also x _{0} itself from the intersection of the lines x = x _{0} and y = x.
Symmetry. Verifying the second property does not present any problem, since a condition in (1.4.25) has a symmetric form, which is not modified when interchanging n and m. If this condition is fulfilled for couples (n, m), it must also be so for couples (m, n). It is simply a well-known property of the commutativity of the logical disjunction in (1.4.25).
A ∖ B = A Δ(A ∩ B).
A ∪ B ∪ C ∪ D = D ∪ (A ∖ B) ∪ (B ∖ C) ∪ (C ∖ D).
(A ∪ B) ∖ C ⊂ (A ∖ C) ∪ B.
A _{t} = [t, t + 1] × [−t, −t + 1].
A _{t} = [−2t − 2, t] × [t − 1, 2t].
$A_t=\{(x,y)\in \mathbb {R}^2\;\; |\;\; (tx-1)^2+t^2y^2< 1\}$.
B is a hexagon with vertices (0, 0), (1, −1), (2, −1), (2, 0), (1, 1), (0, 1); C = {(0, 0)}.
B is a hexagon with vertices (1, 0), (1, 2), (−4, 2), (−4, 0), (−2, −1), (0, −1); C = [−2, 0] ×{0}.
$B=\mathbb {R}_+\!\times \mathbb {R}$; C = K((1, 0), 1).
$A=\{(x+1)/(|x|+2)\;\; |\;\; x\in \mathbb {R}\}$.
$B=\{2/n-3/m\;\; |\;\; n,m\in \mathbb {N}\}$.
$C=\{x\in \mathbb {R},\;\; |\;\; \left |\, |x-1|-|x-2|\, \right |< 2\}$.
Bounded below and above, $\sup X=1\not \!\,\in X$, $\inf X=-1\not \!\,\in X$.
Bounded below and above, $\sup Y=2\not \!\,\in Y$, $\inf Y=-3\not \!\,\in Y$.
Unbounded.
$\mathcal {R}=\{(x,y)\in \mathbb {R}^2\;\; |\;\; \cos x=\cos y\wedge |x|\leq 3\pi \wedge |y|\leq 3\pi \}$.
$\mathcal {R}=\{(x,y)\in \mathbb {R}^2\;\; |\;\; x(x-1)=y(y-1)\}$.
Certain space X and its subset X _{0} are given. Assume that subsets A and B are $\mathcal {R}$-related, if A ΔB ⊂ X _{0}. Investigate whether such relation is an equivalence relation.
Equivalence relation. Classes: e.g., $\left [0\right ]_{\mathcal {R}}=\{ 0,-2\pi ,2\pi \},\;\left [\pi \right ]_{\mathcal {R}}=\{ \pi ,-\pi ,-3\pi ,3\pi \}$.
Equivalence relation. Classes: $\left [x\right ]_{\mathcal {R}}=\{ x,1-x\}$ for x ≠ 1∕2 and $\left [1/2\right ]_{\mathcal {R}}=\{ 1/2\}$.
Equivalence relation. Each set C ⊂ X ∖ X _{0} defines certain equivalence class.