My textbook says "Let $X$ and $Y$ be sets. We say $X$ and $Y$ have the same cardinality if there is a bijection $f: X \to Y$."   I was wondering why the text does not say "if and only if."  A bijection implies same cardinality, but does cardinality imply bijection?  I would imagine so.
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In some books the word "if" is used in definitions and it is not clear if they actually mean "iff" (i.e "if and only if"). I'd like to know if in mathematical literature in general "if" in definitions means "iff".  For example I am reading "Essential topology" and the following definition is written:     In a topological space $T$, a collection $B$ of open subsets of $T$ is said   to form a basis for the topology on $T$ if every open subset of $T$ can be written as a union of sets in $B$.   Should I assume the converse in such a case? Should I assume that given a basis $B$ for a topological space, every open set can be written as a union of sets in $B$?  This is just an example, I am not asking specifically about this sentence.
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So I answered this question:  I believe my answer is correct, however one thing I couldn't explain fully, and which is bugging me, is why the rationals $\mathbb Q$, integers $\mathbb Z$ and naturals $\mathbb N$ all have cardinality $| \mathbb Q |=| \mathbb Z | = |\mathbb N| = \aleph_0$, when $\mathbb N \subsetneq \mathbb Z \subsetneq \mathbb Q$.  The basic proof of the other question, from my answer, is that for any $r$ in $\mathbb Q$ there is a single function $f(r)$ such that $f(r) \in \mathbb Z$, and the same holds between $\mathbb Z$ and $\mathbb N$. Because there is this 1:1 transformation possible, there must be the same number of numbers in the three sets, because otherwise there would be a number in one set for which the bijection could not produce a number of the other set, and this is not so.  Now, Belgi explained why $1&lt;2$ in the other question by defining the value 0 as the cardinality of the empty set $\emptyset$, 1 as the cardinality of a set of sets containing only the empty set, and 2 as the set of sets containing the empty set and a set containing the empty set, then proceeding as follows:     Now that we have defined the natural numbers we can define when one number   is smaller than another. The definition is $x&lt;y$ is $x\subset y$   and $x\neq y$ .      Clearly $\{\emptyset\}\neq\{\emptyset,\{\emptyset\}\}$ and $\{\emptyset\}\subset\neq\{\emptyset,\{\emptyset\}\}$   So this a proof, by definition, of why $1&lt;2$.   ... however, by the same definition, because $\mathbb N \subsetneq \mathbb Z \subsetneq \mathbb Q$, then $| \mathbb N | &lt; | \mathbb Z | &lt; |\mathbb Q|$ and thus a maximum of one of these quantities can be the ordinal quantity $\aleph_0$.
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I was wondering what the cardinality of the set of all real sequences is. A random search through this site says that it is equal to the cardinality of the real numbers. This is very surprising to me, since the cardinality of all rational sequences is the same as the cardinality of reals, and it seemed fairly intuitive to me that if cardinality of a set $A$ is strictly greater than the cardinality of the set $B$, then cardinality of $A^{\mathbb{N}}$ should be strictly greater than cardinality of $B^{\mathbb{N}}$. It turns out to be false.  Some technical answers have appeared in this forum elsewhere but I do not understand them. As I am not an expert in this topic, could some one explain me in simple terms why this is happening?  Also is the cardinality of all functions from reals to reals also the same as the cardinality of reals?
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Let the set S be infinite, and the set T countably infinite. S and T are both subsets of R. Show that S and S U T have the same cardinality.  I know we can discuss whether S is countable or uncountable, but is it true that, for example like this, countable set's cardinality basically does not count in the union of itself and an uncountable set?
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If $A, B, C$ are sets prove that $(A\setminus B)\setminus C\subseteq A\setminus(B\setminus C)$. Find a description of when it is that we have equality, and give an example where the inclusion is strict.   I'm not sure how to do this one I know that to prove that two sets $X$ and $Y$ are equal, we need to show that if $x$ is any element of $X$, then $x\in Y$, and that if $x$ is any element of $Y$, then $x\in X$: the first of these means that $X\subseteq Y$ and the second that $Y\subseteq X$, so the two together establish that $X=Y$. I'm not sure how to use this for a question and how to finish it.
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I cannot prove this statement so need help. This problem is one of exercises right after the chapter about Hausdorff's maximal principle and Zorn's Lemma. Thus, you cannot use the concept of cardinal number. I found the exact same question on this website, but the proof is done by using the concept of cardinal number. Please make sure that this question is at right after introducing the axiom of choice and Zorn's lemma.  Let $A$ be any set with more than one element. Prove that there exists a bijective function $f:A \to A$ such that $f(x)\neq x$ for every element $x$ in $A$.
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This seems to be a common result. I've been trying to follow the bijective proof of it, which can be found easily online, but the explanations go over my head. It would be wonderful if you could give me an understandable explanation of the proof and let me know how I'd go about finding such a bijection.
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I Have a question. Given only the definition of equality of two sets, how can we prove that there is one and only one empty set. I mean by equality of two sets the following: $$A=B \iff  \forall x (x \in A \iff x \in B)$$
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I am wondering how to show the following proposition  If A, B, and C are set and $A \subset  B ,B \subset C \rightarrow A \subset C$  My proof   Let x be any integer.  If $x\in A$ then $x\in B$ because all element of A are in B.  If $x \in B \rightarrow x\in C$ because all element of B are in C.  Since B contains A and C contains B, then C contains A logically.  the end
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I came across this long proof on this site:    But I would like to know whether my direction can work.  Say we want to find the cardinality of all equivalence relations in $\mathbb{N}$. Since it is a subset of all relations in $\mathbb{N}$, I conclude it has a cardinality smaller or equal to $\aleph$. Now, define an injective function from $P(\mathbb{N})$ to the set of equivalence relations by matching each subset of $\mathbb{N}$ with the identity relation (which is an equivalence relation in $\mathbb{N}$.  Therefore the cardinality of all equivalence relations in $\mathbb{N}$ is greater or equal to $\aleph$ and using CSB we get the desired result.  Seems legit?
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Is the cardinality of the Cartesian product of two equinumerous infinite sets the same as the cardinality of any one of the sets? I couldn't find this explicitly stated in any handout or text. This certainly seems to be true from the examples I have seen:  The Cartesian product of two infinitely countable sets is again infinitely countable. The Cartesian product of two sets with cardinality of continuum again has cardinality of continuum.
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This is an exercise of an assignment I have:     Suppose $A$ and $B$ are finite sets and $f\colon A\to B$ is surjective. Is it   true that the relation “$|A| &lt; |B|$” is a sufficient condition for   claiming that $f$ is a bijection? Justify your answer.   And this is my answer:     No. In fact, if $|A| &lt; |B|$, then there should exist at least $1$ element   of $A$ that points to more than $1$ element of $B$, since all elements of $B$   must be pointed (surjective), but this is not a function,   because the same element of $A$ point to different elements of $B$.   Is my answer correct? I don't know if the question actually is asking for a proof or what, and if my answer is a proof, if correct.
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Let $A$ be a set with card($A$)=$a$. What is the cardinal number of the set of countably infinite subsets of $A$?   I see that this problem is equivalent to finding the cardinal number of the set of injective functions from $\mathbb{N}\rightarrow{A}$. I also know that the cardinal number of the set of bijections from $A\rightarrow{A}$ is $a^{a}$.   Hints and general heurisitcs would be greatly appreciated.
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To familiarize myself with axiomatic set theory, I am reading Kenneth Kunen's  that presents ZF set theory. I haven't gotten really far since I am stuck at the axiom of Extensionality, stated as follows:  $$\forall x,y \; [\forall z(z \in x \leftrightarrow z \in y) \rightarrow x = y]$$  As far as I understood it, the purpose of this axiom is to state that every two sets that have exactly the same members are the same set. In terms of the above formula, when antecedent is true, the consequent needs to be true. What confuses me here is the case when the antecedent is false. Then, the consequent can be false or true, i.e., we can't say much about the consequent. But, don't we really want to say that in that case those two sets are not equal? More precisely, shouldn't we use equivalence instead of implication? If not, what is the reason?
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Let $A$, $B$ be sets and denote by $A\leq^* B$ an assertion that if $A$ is non-empty, then there's a surjection from $B$ onto $A$. This might be understood as a dual definition of cardinality. (Typically we use injections to capture the notion of A not being bigger than B. The $\leq^*$ notation is to be read as B not being smaller than A.)  Is this relation trichotomous without using the axiom of choice? Meaning for every two sets $A$ and $B$, it holds $A\leq^* B$ or $B\leq^* A$. (I would be surprised). Would trichotomy imply some interesting extra-ZF axiom?  More broadly, is there anything to be said about $\leq^*$? For example how does it relate to the usual comparisons of cardinality by injections?  I suspect there's no clear answer, since it is somehow reminiscent of Partition Principle ($A\leq^* B \implies A\leq B$) and Weak Partition Principle ($A\leq^* B \implies B\not&lt; A $) which both are independent of ZF.
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This proof seems odd to me. I have come to the conclusion that I will use induction. I would like to see a smoother way or just some improvements on my technique.     Let $f:A\rightarrow B$ be a function between two finite sets of equal cardinality. Show that $f$ is surjective if and only if it  is injective.   To start, I will show that a surjection implies an injection using induction. I will dismiss the cases that both sets are empty or contain one element as being trivial (essentially vacuously true).   Assume $|A| \geq 2$, $|B| \geq 2$, $|A| = n = |B|$, and $f:A \rightarrow B$ is a surjection. For the base case, let $n = 2$.   There are two elements in both $A$ and $B$. Due to surjection, every element $b \in B$ must be mapped to, through $f$, by at least one element $a \in A$. If each of the two elements in $B$ were mapped to by the same element in $A$, the definition of function would be violated. Therefore, they are mapped to by unique elements in $A$. Thus, for $f(p), f(q) \in B$, if $f(p) = f(q)$, it must be true that $p = q$ so $f$ is injective.  Now assume that the surjection implies an injection for $n \geq 2$. We must show this to be true for $|A| = n + 1 = |B|$. Since it is true for $|A| = n = |B|$, the $n + 1$ case represents the addition of one new element to both $A$ and $B$. The new element in $B$ cannot be mapped to any other element in $A$ except for the new one. If mapped to by an old one, the definition of function would be violated. It must be mapped to by something since $f$ is surjective, hence it must be the new element. Finally, the new element in $A$ cannot be mapped to an old element in $B$ because it is unique and the previous $B$ was shown to be injective.  $$\blacksquare$$  This is a very wordy and awkward proof in my opinion. I have been out of proofs for a long time. I would like to see one that is more clear or seek validation if there isn't. I know that I have only completed half of the proof and have yet to go the other way.
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What is the simplest way of proving (to a non-mathematician) that the power set of the set of natural numbers has the same cardinality as the set of the real numbers, i.e. how to construct a bijection from $\mathcal{P}(\mathbb{N})$ to $\mathbb{R}$?