# Tagged Questions

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### Elementary set problem

Let $A$ be a set and $f$ a function $f:A \to A$. A set $B$ is called "solid" in $A$ if and only if $f(B) \subseteq B$. Prove that a set $A$ is finite if and only if there is a function $f:A\to A$ ...
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### Set notation and mappings question

Good evening. I have a question. Suppose I have two sets, $A=\{1,2,3,4\}$ and $B=\{5,6\}$. I want to write the notation for a function that takes each element in $A$ and assigns to it a value in $B$. ...
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### Help explain the set being constructed in this Cantor-Schroder-Berstein proof

The Cantor-Schroder-Bernstein theorem states that: Suppose $A$ and $B$ are sets. If $|A|\le |B|$ and $|B|\le |A|$, then $|A|=|B|$ Proof: So, $|A|\le|B|$ implies we can choose an injection ...
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### Notion of surjective functions

I have some issues with fully understanding the concept of surjection. For the proof of $A_{m}$ being a countable set for each m $\in$ N, then the union $A = \bigcup_{m=1}^{\infty} A_{m}$ is ...
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### Surjectivity of a piecewise function $f:(-1,1)\to \mathbb R$

Function $f$ is defined as $f: (-1, 1) \to \mathbb{R}$. $$f(x) = \begin{cases} -x/(x-1),&x\geq 0 \\ x/(x+1),&x \leq 0 \end{cases}$$ Let $y \in \mathbb{R}$. How would I prove that there ...
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### Explanation of the formula $f^{-1}(Y)=\{x \in A |f(x) \in Y\}$ for the preimage of a set

So I found a Definition in the book that goes like this to find the pre-image of a set: $$f^{-1}(Y)=\{x \in A |f(x) \in Y\}$$ Example of the theorem being used: Let $A = \{1,2,3,4,5,6\}$ and ...
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### Relations on set A and conditions of existence of descending chains

I am reading through the Elements of Set theory by Herbert Enderton and even though I have passed this exercise long ago,the more I look at my solution the less I believe it. Problem goes like this: ...
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### Proof that there is a bijection, if there are injective maps in both directions

Let $A$ and $B$ be two sets. Let $f:A\to B$ be injective such that $Im(f) \subsetneq B$. Let $g:B\to A$ be injective such that $Im(g) \subsetneq A$. Obviously $A$ and $B$ are not finite sets. Can ...
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### Proof for Surjections

I'm reading through Basic Algebra I (which I enjoy so far. Thoughts on this for self-studying?) and am having a difficult time proving surjection. I believe I understand the concept, but when it comes ...
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### How to deduce number of unordered distinct pairs using set operations and bijections

In (b) of the example, we are ask to calculate the number of ordered pairs of distinct positive integers. I like the first method's answer (using bijections, set operations) because it clearly shows ...
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### Two definitions of functions

In literature on logic and set theory, there seem to be two different definitions of functions, one more general than the other. First of all, a function $f\colon X\to Y$ consists of three element ...
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### Showing that $A\rightarrowtail A \times \{x\}$ is a bijection

$A\rightarrowtail A \times \{x\}$ where $A$ is any set and $\{x\}$ is an arbitrary one-object set. How would I show the following is a bijection ( one to one and onto)? I know if I turn it into a ...
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### Composition of injections (proof)

I'm trying to prove that composition of injections is an injection. I want to know if this is a good proof: Composition of injections is an injection. Let $f:S_1\rightarrow S_2$ and ...
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### What would make a function reflexive, transitive, and/or symmetric?

A binary relation $R$ is a subset of the Cartesian product between two sets $X$ and $Y$, containing a set of ordered pairs $\{(x,y) : x \in X, y \in Y\}$. $R$ is a function if each element of $X$ is ...
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### Suppose $f: X \rightarrow Y$, and is one-to-one, and let $A \subseteq X$, prove that $f^{-1}[f[A]] = A$. [duplicate]

Suppose $f: X \rightarrow Y$, and is one-to-one, and let $A \subseteq X$, prove that $f^{-1}[f[A]] = A$. EDIT: Actually, this identity should hold even if $f$ is not one-to-one (injective), right? ...
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### Understanding the difference between relations and functions.

$R=\{(1,2),(1,3)\}$ is a relation but not function. The logic for this is that if the first element of every ordered pair must remain different, then it is said to be function. Otherwise, it's just ...
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### Does something that is injective, surjective or bijective imply that it is a function?

As the title says. Sorry it seems like a silly question but it's something I've been wondering because it seems like sometimes the word "function" is omitted, but other times it is included
Assume $f:A\to A'$ is a function, $B\subset A'$, $C\subset A'$, and $f^{-1}(B\cap C)=\emptyset$ How can we see that $f^{-1}(B)\cap f^{-1}(C)=\emptyset$?