# Tag Info

## New answers tagged elementary-set-theory

1

When Velleman writes $$I_n = \{i \in \mathbb Z^+ |\ i \le n\}$$ he is using set-builder notation to describe a certain set: The set of positive integers, ($i \in \mathbb Z^+$) that are less than or equal to n ($i \le n$). For example, $I_3 = \{1,2,3\}$ because 1, 2, and 3 are the positive numbers less than or equal to 3. Without using the notation, one ...

2

Suppose that $A$ is a set of physical objects in the room with you, and you have a bijection $f:I_{42}\to A$. You can point to $f(1)$, which is an element of $A$, and say "one" to yourself. You can then point to $f(2)$, which is a different element of $A$, and say "two" to yourself. Continuing, eventually you point to $f(42)$, say "forty-two", and you're ...

1

So $I_n=\{1,2,3,\dots,n\}$. For instance, $I_5=\{1,2,3,4,5\}$. Each of these $I_n$ sets give a standard-sized finite set. In order for a set to be finite it must be equinumerous with one of these standard finite sets (or it must be empty).

1

$D \notin (X-Y)$ if and only if $\neg (D \in X \wedge D \notin Y)$. Because of De Morgan's laws $$\neg (D \in X \wedge D \notin Y) \iff D \notin X \vee D \in Y$$ If we suppose that $D\in X$, than $D \in Y$ must be true.

1

Since you already answered that we are assuming $D \in X$, then suppose $D \not \in (X - Y)$. Note that $X - Y = X \cap Y^{c}$. So we have $D \not \in X - Y \implies D \not \in X \cap Y^{c}$. But $D \in X$, and $X = (X \cap Y) \cup (X \cap Y^{c})$, so if $D \in X$, then $D$ must be in one of the two sets $(X \cap Y)$ or $(X \cap Y^{c})$. Since we ...

1

If $D \in X$ and $D \not\in Y$, then by definition $D \in (X-Y)$. But since we know that $D \not\in (X-Y)$ we must have either $D \not\in X$ or $D \in Y$. Since you said you know that $D \in X$, then we are left with $D \in Y$. The statement $D \not\in (X-Y) \iff D \in Y$ is like saying "Since the cookie jar is not in any of the other rooms in the ...

6

$$d \notin (X-Y) \iff \lnot(d \in X \land d\notin Y) \iff (d\notin X \lor d\in Y)$$ Since we know that $d\in X$, we know that $$\lnot \lnot (d \in X) \iff \lnot(d\notin X)$$ which, together with the first premise, it follows that $d \in Y$.

3

This is not necessarily true, if $D\notin X$. Of course, in the context from which you take it, you already assume that $D\in X$, so it is necessarily the case that $D\in Y$.

2


1

If $x\in F$, then $x$ is element of one of the sets that are forming the union $F$, therefore $x\in \cap_{k = n}^\infty E_k$ for at least one $n\in\mathbb{N}$. Because $x$ is the element of the intersecion $\cap_{k = n}^\infty E_k$, $x$ is element of every $E_k$ where $k\geq n$. One can conclude that $x$ may be "missing" only in the first $n-1$ sets, which ...

0

As to a): $x \in F$ iff there exists some $n(x)$ such that for all $k \ge n(x)$, $x \in E_k$. So $x$ is in eventually all (except maybe the first $n(x)$ many) $E_k$... As to b): $x \in G$ iff for every $n$ there exists some $j \ge n$ such that $x \in E_j$, so we can always find "new" indices $j$ such that $E_j$ contains $x$.

2

A non principal ultrafilter on the set of natural numbers is a tail set but it is far from Borel. What you are missing is that you need to make use of the fact that your set is Borel (or Lebesgue measurable) so that you can apply something like Lebesgue density theorem.

1


0

I had to define an n-tuple his way: I had to come up with a property $P(x, y)$ and state that an object $a$ is an n-tuple if it satisfies the property $P(a, n)$. Then I had to prove 1) and 2) parts as theorems. So parts 1) and 2) are not the way to define it, they have to follow from your definition. Look at the whole hint the author provides for this ...

3

The reason would depend on the particular argument you give but it all boils down to the fact that (with the axiom of choice) one can construct very bizarre sets of real numbers. You also have to be more precise. There are measures on the set of all subsets of $\mathbb R$, e.g., counting measure. What does not exist is a translation invariant extension of ...

1

Yes. $A\cap B\cap C\cap D=A\cap B\cap(C\cap D)$ is the intersection of the three sets $A$ and $B$ and $C\cap D$, so it is empty.

2

$A \cap B \cap C \cap D \subseteq A \cap B \cap C = \emptyset$. Thus $A \cap B \cap C \cap D = \emptyset$.

2

Yes, if $A$, $B$, $C$, and $D$ are sets such that the intersection of any three is empty, then because intersection is associative, we have that $A \cap B \cap C \cap D = (A \cap B \cap C) \cap D = \emptyset \cap D = \emptyset$. Just in case you aren't sure what associative means, it means that we can intersect in any order. So, in the case of three sets, ...

2

Yes. Let's suppose that $A \cap B \cap C = \emptyset$. Then $A \cap B \cap C \cap D = (A \cap B \cap C) \cap D = \emptyset \cap D = \emptyset$.

0

Let $S_{2n+1}$ denote the set of string of length exactly $2n+1$. Since this set is finite, there is a mapping $f_n:S_{2n+1} \to |S_{2n+1}|$ (where |S| denotes the cardinality of $S$)which lists all elements in $S_{2n+1}$. Now consider the function $f:\cup_{n \in \mathbb{N}}S_{2n+1} \to \mathbb{N}$ defined by ...

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