Problem 1 asks to show that the relation defined as follows is antisymmetric: Given a set the relation is defined on the subsets of by setting for iff where denotes set-theoretic difference of sets.

To show that is antisymmetric, we need to show that whenever are such that and then Suppose satisfy these assumptions. We need to show that they are equal as sets, i.e., that they have the same elements.

By definition, holds iff and holds iff Recall that if are sets, then It follows that iff every element of is also an element of and that iff every element of is also an element of But these two facts together mean precisely that and have the same elements, i.e., that as we needed to show.

Problem 2(a) of quiz 6 asks to consider the relation defined on by setting for iff and to show that is an equivalence relation. This means that is reflexive, symmetric, and transitive. Problem 2(a) of quiz 5 asks to show one of these properties.

To show that is reflexive, we need to show that for any we have that i.e., that But is certainly divisible by 5.

To show that is symmetric, we need to show that for any if it is the case that then it is also the case that Suppose then that This means that i.e., there is an integer such that But then showing that also i.e.,

To show that is transitive, we need to show that if and both and hold, then also holds. But if and then and But then it is certainly the case that Since this proves that, indeed or as we needed to show.

(If one feels the need to be somewhat more strict: That means that there is an integer such that Similarly, means that there is an integer such that But then showing that there is an integer such that namely, we can take )

Problem 2(b) asks to find all natural numbers such that where is as defined for problem 2(a).

That means exactly the same that Since and an integer is a multiple of 5 iff it ends in 5 or 0, we need to find all natural numbers such that ends in or Since and is even for all naturals we actually need to find all natural numbers such that ends in 8. For this, we only need to examine the last digit of the numbers and we find that these last digits form the sequence which is periodic, repeating itself each 4. This means that the numbers we are looking for are precisely i.e., the natural numbers of the form with

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I was working with one of students(Josue Gomez) in your class.
From problem 2 (b,)
I guess 5 divides 0 since 5*0 = 0. so need to say n is the form 4k + 3 where k is greater than or equal to 0.
I understand 0 is not natural number. I guess k can be 0 also.

Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here. What would be an explicit example of a number $r$ with this property? Short of an explicit example, are there any references addressing this question? A natural approach would […]

Suppose $M$ is an inner model (of $\mathsf{ZF}$) with the same reals as $V$, and let $A\subseteq \mathbb R$ be a set of reals in $M$. Suppose further that $A$ is determined in $M$. Under these assumptions, $A$ is also determined in $V$. The point is that since winning strategies are coded by reals, and any possible run of the game for $A$ is coded by a real, […]

Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.) Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so, provably in $\mathsf{ZFC}$, the relevant […]

Please send an email to mathrev@ams.org, explaining the issue. (This is our all-purpose email address; any mistakes you discover, not just regarding references, you can let us know there.) Give us some time, I promise we'll get to it. However, if it seems as if the request somehow fell through the cracks, you can always contact one of your friendly edit […]

The characterization mentioned by Mohammad in his answer really dates back to Lev Bukovský in the early 70s, and, as Ralf and Fabiana recognize in their note, has nothing to do with $L$ or with reals (in their note, they indicate that after proving their result, they realized they had essentially rediscovered Bukovský's theorem). See MR0332477 (48 #1080 […]

The paper MR1029909 (91b:03090). Mekler, Alan H.; Shelah, Saharon. The consistency strength of "every stationary set reflects". Israel J. Math. 67 (1989), no. 3, 353–366, that you mention in the question actually provides the relevant references and explains the key idea of the argument. Note first that $\kappa$ is assumed regular. They refer to MR […]

Start with Conway's base 13 function $c $ (whose range on any interval is all of $\mathbb R $), which is everywhere discontinuous, and note that if $f $ only takes values $0$ and $1$, then $c+f $ is again everywhere discontinuous (since its range on any interval is unbounded). Now note that there are $2^\mathfrak c $ such functions $f $: the characteris […]

Yes, there are such sets. To describe an example, let's start with simpler tasks. If we just want $P\ne\emptyset$ with $P^1=\emptyset$, take $P$ to be a singleton. If we want $P^1\ne\emptyset$ and $P^2=\emptyset$, take $P$ to be a strictly increasing sequence together with its limit $a$. Then $P^1=\{a\}$. If we want $P^2\ne\emptyset$ and $P^3=\emptyset$ […]

The result was proved by Kenneth J. Falconer. The reference is MR0629593 (82m:05031). Falconer, K. J. The realization of distances in measurable subsets covering $R^n$. J. Combin. Theory Ser. A 31 (1981), no. 2, 184–189. The argument is relatively simple, you need a decent understanding of the Lebesgue density theorem, and some basic properties of Lebesgue m […]

No, not even $\mathsf{DC}$ suffices for this. Here, $\mathsf{DC}$ is the axiom of dependent choice, which is strictly stronger than countable choice. For instance, it is a theorem of $\mathsf{ZF}$ that for any set $X$, the set $\mathcal{WO}(X)$ of subsets of $X$ that are well-orderable has size strictly larger than the size of $X$. This is a result of Tarski […]

Dr. Caicedo,

I was working with one of students(Josue Gomez) in your class.

From problem 2 (b,)

I guess 5 divides 0 since 5*0 = 0. so need to say n is the form 4k + 3 where k is greater than or equal to 0.

I understand 0 is not natural number. I guess k can be 0 also.

Thanks for your lecture!

Ei

The

natural numbersbegin with 0. That is the definition we are using. The integers that are larger than 0 are, naturally, thepositive integers.Oh thank you!