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Yes, it's true - take $1, 1 + 1, 1 + 1 + 1, \dots$. Indeed, addition obeys the order, so this must be a strictly increasing sequence; trichotomy then guarantees that none of these are equal.

Similarly, what's wrong with $1 \le 1 \le 1 \le \dots$ (or where $\le$ is replaced by $\ge$)? If you're happy with this, then you should be happy with the sets.

I'm happy enough with that. My own answer would go: it is enough to show that any two infinite subsets of $\mathbb{N}$ have the same cardinality, since $\mathbb{N}$ and $\mathbb{Z}$ are in bijection. …

One way is to take $(a, b)$ to $$2^{\vert a \vert} 3^{\vert b \vert} 5^{\overline{\text{sgn}}(a)} 7^{\overline{\text{sgn}}(b)}$$ where $\overline{\text{sgn}}(a)$ is $1$ if $a>0$, $0$ if $a \leq 0$. …

Authors differ a bit on their definitions here. A "proper subset" of $A$ is always(?) not $A$ itself, but $\emptyset$ may (albeit rarely) be referred to as "proper". The idea is that when we select a …

You need to find a set $A$, none of whose members are subsets of $A$. There's lots of those! The simplest possible example would be a set of one element; can you find one which works? Another hint: t …

Note that $\emptyset$ is an ordinal. So the intersection $\bigcap \mathcal{O}$ is an intersection of a family which contains the empty set, so is empty.

$\mathcal{P}(\mathbb{N}) \times \mathbb{N}$ is the set of all 2-tuples such that the first element is a set of naturals and the second is a natural. So, for example, it contains $(\emptyset, 2)$. I …

The expression is $(X^C \cup Y^C) \cap Y^C$, which simplifies to $Y^C$. Indeed, $z \in (X \cap Y)^C$ if and only if $z \not \in X \cap Y$, if and only if it is either not in $X$ or not in $Y$. That i …

You are mistaken. $\{x \in \emptyset \mid p(x) \}$ is still a set; it's just empty. It has no elements, because as you point out, if it had an element then that element would lie in the empty set (a c …

It depends how you implement the ordered pair, and how you implement the natural number. The usual implementations are: the Kuratowski definition of the ordered pair: $(a,b) = \{ \{a\}, \{a, b\}\}$ …

There is a much easier way to do the first direction: show that both $A \cup B$ and $A \cap B$ are equal to something else, rather than showing them to be equal to each other. In fact, I don't see a w …

This answer is incomplete, but it at least makes the Schröder-Bernstein a bit nicer. Firstly, $[0,1)$ bijects with $\mathbb{R}$, by the following bijections: $h: (0, 1) \to \mathbb{R}$ by $x \mapsto …

If you've shown $X$ contains a countable subset $A = \{ a_1, a_2, \dots \}$, then you can throw away $a_1$ in the following way. By the inclusion map, $X \setminus \{ a_1 \}$ clearly injects into $X …

Your question appears to be: how does the "interlacing" work in an existing answer? To enumerate the disjoint union $A + B$ given an enumeration of $A$ and an enumeration of $B$, simply hop between th …