$$f(A\cup B)=f(A)\cup f(B)$$  My attempt:  Suppose that $f(A\cup B) \neq f(A) \cup f(B)$, then $f(A\cup B)$ has a variable $x$ that is neither in $A$ nor $B$; however, $A\cup B$ implies that $x$ must me in either $A$ or $B$, thus we have a contradiction.  Yeah... it's terrible, but I have no intuition as to how to prove this in any other way. Also, the proofs I saw in the lectures were sometimes as unconvincing as this one I just came up with.  It's like they're saying this is this because it is so. How do you develop the skills and intuition to solve these kind of problems?  Also, you have to know it is true before proving it. How do you even know it is true just by looking at it?
14606
$f(S \cup T) = f(S) \cup f(T)$  $f(S)$ encompasses all $x$ that is in $S$ $f(T)$ encompasses all $x$ that is in $T$  Thus the domain being the same, both the LHS and RHS map to the same $y$, since the function $f$ is the same for both.  Can you post the solution?
99112
I had the following two problems:     Find a counterexample for $f_*(A \cap B) \supseteq f_*(A) \cap f_*(B)$ and $ f_*(A-B) \subseteq f_*(A) -f_*(B).$ Where $f_*(X)$ is the image of $X$ under $f$ for some function $f:A \rightarrow B $ and some subset $X \subseteq A$.   I came with the following answer:  Let $ f: \mathbb{Z} \rightarrow \mathbb{Z} $, where $f: x\mapsto 0$; let $A=\{1,2\}$, and $B=\{3,4\}$.     For $f_*(A \cap B) \supseteq f_*(A) \cap f_*(B)$, $$ f_*(A \cap B) = f_*\{ \varnothing\} = \varnothing \hspace{1cm} \text{and} \hspace{1cm} f_*(A) \cap f_*(B)=\{0\} \cap \{0\} = \{0\}. $$ It follows that $ \varnothing \nsupseteq \{0\} $. For $ f_*(A-B) \subseteq f_*(A) -f_*(B)$, $$f_*(A-B)=f_*(\{1,2\})=\{0\} \hspace{1cm} \text{and} \hspace{1cm} f_*(A) - f_*(B) = \{0\} - \{0\} = \varnothing$$ It follows that $ \{0\}  \nsubseteq \varnothing $.   Now (assuming my answer is correct), let's say I want to find more counterexamples or maybe a more general answer (e.g. all the conditions under which the inclusion fails). Other than by trial and error or just intuition, how would I do that?   I know this is a broad question, so I'm specifically asking for methods or advice on how to build counterexamples in elementary set theory (for example, tips on where to begin or rules that apply in general that may be unobvious).
35262
Given two sets $A$ and $B$, what is the difference between $A + B$ and $A \cup B$? For example, if $A = \left\{ a, b, c \right\}$ and $B = \left\{ d, e, f \right\}$, what are $A + B$ and $A \cup B$, respectively?
63529
I want to prove this statement:  $$(A_1 \cup A_2)^c  =  {A_1}^c \cup {A_2}^c$$  where the $c$ means the complement.  Any help would be greatly appreciated.
118585
Theorem: If $X$ is a set, then $X$ is not equivalent to its power set.  Proof: suppose for a contradiction that $f:X\to P(X)$ is a bijection. Define $B:=\{x \in X, x\not\in f(x)\}$. Because $f$ is surjective (onto), there is a $b$ in $X$ with $B=f(b)$. Now $b$ in $B$ would imply $b$ in $B=f(b)$. By definition of $B$, this would mean that $b$ is not an element of $f(b)=B$. Thus we infer $b$ is not an element of $B$. But then $b$ is not an element of $B=f(b)$, which by definition of $B$ forces $b$ in $B$, a contradiction.  I understood this proof until he reached the part that says "Now $b$ in $B$ would imply $b$ in $B=f(b)$..."  This seems problematic to me because the assumption $b$ in $B$ seems unwarranted. Since $B$ is a subset of $X$ it makes sense if we had $b$ in $B$ and then arrived at $b$ in $X$ but the reverse does not seem like a valid step and I'm left wondering if the resulting contradiction that arised came simply from the assumption $b$ is in $B$ and therefore only means that $b$ is not not an element of $B$, And that this has no bearing on the bijection part of the argument. In fact I dont even see how this connects to the part of the argument that used the onto property of bijectiveness.
94485
I'm pretty sure I've asked way too many questions on sets already(still, a big thanks to everyone who helped) and even though I'm starting to see what's going on there are still some issues, for example this question:  Prove that $\bigcup P(X)=X$, where $P(X)$ is the power set for the set $X$  So, what I can do is the following:  $a \in \bigcup P(X) \implies \exists b \subset X$ s.t. $a\in b \implies a \in X$ and for the other direction I can take $b=X$.  However, I'm not sure whether this is correct, and in particular, the bit with $b \subset X, a\in b \implies a \in X$  If I take $X=\{\{1\},\{2\}\}$, then $P(X)=\{\emptyset,\{1\},\{2\},\{\{1\},\{2\}\}\}$, hence $\bigcup P(X)=\{1,2,\{1\},\{2\}\}$ which is not equal to $X$ by ZF$1$  Where's the problem?  Thanks
56462
Show that if $A$ and $B$ are sets, then $(A\cap B) \cup (A\cap \overline{B})=A$.  So I have to show that $(A\cap B) \cup (A\cap \overline{B})\subseteq A$ and that $A \subseteq(A\cap B) \cup (A\cap \overline{B})$.  Lets begin with the first one:  If $x \in (A \cap B)$ it means $x \in A \wedge x \in B$.  If $x \in (A \cap \overline{B})$ it means $x \in A \wedge x \in \overline{B}$.  And the second one:  If $x \in A$ it means $x \in (A \cap B)$.  But here after I am confused.
3197
How do I prove or disprove $X \times (Y \cup Z)= (X \times Y) \cup(X\times Z)$  for all sets $X$, $Y$, and $Z$? I'm lost on the steps here.
85448
What I've done so far is stated that    $(A \cup A) \cap (A \cup B)$  by distribution $A\cap (A\cup B)$        by 1, definition of $\cup$ $A\cap A$              by 2, definition of $\cup$ $A$                   by 3, definition of $\cap$   I'm not sure if I need to state that $A\cup B = \{x\mid x\in A\lor x\in B\}$ somewhere in there.  I'd really appreciate some help on this one. Thank you!
88648
I am trying to define a proof for  $A \cup B = B$ if and only if $A \subset B$  I started with:  $$A \cup B$$ $$x \in A \lor x \in B$$   but I am unsure on how to continue.
79978
Is this proof correct? To prove $F(A)\cap F(B)\subseteq F(A\cap B)  $ for all functions $F$.  Let any number $y\in F(A)\cap F(B)$. We want to show $y\in F(A\cap B).$ Therefore, $y\in F(A)$ and $y\in F(B)$, by definition of intersection. By definition of inverse, $y=F(x)$ for some $x\in A$ and $x\in B$ And so, $y=F(x)$ for some $x\in A\cap B$ And therefore, $y\in F(A\cap B)$  I have a gut feeling deep down that something is wrong. Can anyone help me pinpoint the mistake? I am not sure why am I having so much problems with functions. Am I not thinking in the right direction? Sources : 2nd Ed, P219 9.60 = 3rd Ed, P235 9.12, 9.29 - Mathematical Proofs, by Gary Chartrand, P214 Theorem 12.4 - Book of Proof, by Richard Hammack, P257-258 - How to Prove It, by D Velleman.
1790
Let $f$ be a real valued and continuous function over $[a,b]$ with $a &lt;b$ such that $f(a)=f(b)$ , My question here is :     Question (claim): How to show that : there  exists  $c \in [a, \frac{a+b}{2} ]$ such that : $f(c)=f(c+\frac{b-a}{2})$.   Note: I have tried intermediate values but I didn't succeed.
3279
I have to resit a calculus exam and for some reason set proofs were never my best friend...  Anyway, on a practice exam I encountered the following proof:  $$A\cap(B\cup C) = (A\cap B)\cup(A\cap C)$$  When I draw a Venn-diagram it seems quite obvious but I couldn't manage to write the proof down properly.  If someone could help me, that'd be great!
50057
Two functions $f:A\rightarrow B$ and $g:B\rightarrow A$. Can someone give me an example where $g\circ f=id_A$ but $f\circ g\neq id_B$?
60912
We were ask to prove that $|A \cup B| = |A| + |B| - |A \cap B|‫‪$. It was easy to prove it using a Venn diagram, but I think we might be expected to do if more formally. Is there a formal way?
111724
The Problem:   Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $$f(x + y) \leq yf(x) + f(f(x))$$ for all real numbers $x$ and $y$.  Prove that $f(x) = 0$ for all $x ≤ 0$.  (IMO $2011$ , Pr: $3$)  Here is the solution:  Before we prove that $f (x) = 0$ for all $x ≤ 0$, we will prove that $f (0) = 0.$ $\color{black}{\large\text{Case} \thinspace  1:}$ $f(0)&gt;0$ Let, $\varepsilon &gt;\dfrac{f(f(0))}{f(0)}$ and $x&lt;- \dfrac{\varepsilon+f(f(0))}{f(0)}$  we have, from $f(x)\leq xf(0)+f(f(0))$,  $f(x)&lt;-\varepsilon.$ Then from, $f(f(x))\leq f(x)f(0)+f(f(0))$ we have $f(f(x))&lt;0$. Finally,  since $0&lt;f(0)\leq f(f(0))$ ,we get $f(0)\leq-xf(x)+f(f(x))&lt;0$ which imply $f(0)&lt;0$, which gives a contradiction. $\color{black}{\large\text{Case} \thinspace  2:}$ $-\alpha \leq f(0)&lt;0$, where $\alpha\in\mathbb{R^+}.$ From $f(x + y) \leq yf(x) + f(f(x))$ we have: $\begin{cases} f(x)\leq f(f(x)) \\ f(x) \leq xf(0)+ f(f(0)) \end{cases} \Longrightarrow f(x)\leq f(x)f(0)+f(f(0)) \Longrightarrow f(x)(1-f(0))\leq f(f(0))$. Then applying $x=f(0)$, we get $f(f(0))\leq 0$, which imply $f(x)\leq 0$, which gives $f(f(x))\leq 0$. In this case, we have $f(x)&lt;0.$ Because, if $f(x)=0$ at least for some $x$, from $f(x)\leq f(f(x))$, we get $f(0)\geq 0$, which gives a contradiction. Then let, $x&lt;-\sqrt {\alpha}$,  from  $f(x + y) \leq yf(x) + f(f(x))$  we have: $$\begin{align} f(z)\leq(z-x)f(x)+f(f(x)) \Longrightarrow f(f(x)) \leq (f(x)-x)f(x)+f(f(x))\Longrightarrow f(x)(f(x)-x)\geq 0 \Longrightarrow f(x) \leq x &lt;-\sqrt {\alpha} \Longrightarrow f(x)&lt; -\sqrt{\alpha}.\end{align}$$ Finally, from $f(0)\leq-xf(x)+f(f(x))$ we get $f(0)&lt;-\alpha$, which gives a contradiction. So, we deduce that $f(0)=0$. Then, applying $f(0)=0$, from  $f(0)\leq-xf(x)+f(f(x))$ and $f(x) \leq xf(0)+ f(f(0))$ we have, $$\begin{cases} xf(x)-f(f(x)) \leq 0 \\ f(x) \leq 0 \\ f(f(x)) \leq 0 \end{cases} \Longrightarrow  \begin{cases} xf(x)\leq0 \\ f(x)\leq 0 \end{cases} \Longrightarrow \begin{cases} xf(x)\leq0, x\in\mathbb {R} \\ xf(x)\geq 0, x\leq 0 \end{cases} \Longrightarrow xf(x)=0, x\leq 0 \Longrightarrow \color{blue} {\boxed{ f(x)=0, \text{for all} \thinspace x\leq 0.}}$$ End of the Proof.  Can you verify this solution? Are there any mistakes, gaps, etc.? Thank you very much!
105971
I have been learning set theory from scratch again and I have a question about the distributive property for sets. It seems the only proof I found of it actually USES the very same distributive property while proving it. This seems circular to me, could someone explain why it is allowed?  This is what I am trying to prove. $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$  I got this proof from a quora answer.  Let $x \in A \cup (B \cap C)$. If $x \in A \cup (B \cap C)$ then $x$ is either in $A$ or in ($B \cap C$).  $x \in A$ or $x \in (B \cap C)$  $x \in A$ or ($x \in B$ and $x \in C$)  HERE!!! How does one get from the statement above to the one below without actually using the distributive property itself?  ($x \in A$ or $x \in B$) and ($x \in A$ or $x \in C$)  $x \in (A \cup B)$ and $x \in (A \cup C)$  $x \in (A ∪ B) \cap (A ∪ C)$  $x \in A \cup (B \cap C) \implies x \in (A \cup B) \cap (A \cup C)$.  Thanks a lot!!