I have a subset $B = \{(x,y) : x^2 + y^2 \leq 1 \} \subseteq \mathbb{R}^2$.  I am trying to prove that function   $g(x,y) = (\frac{x^2 + y^2}{x^2 + y^2 + 1}x, \frac{x^2 + y^2}{x^2 + y^2 + 1}y)$  is an injective function $g: \mathbb{R}^2 \rightarrow B$.  I attempted to prove this by directly applying the definition of injection.  Suppose we have $(a,b),(c,d) \in \mathbb{R}^2$ such that   $g(a,b) = g(c,d)$.   If I can prove $g(a,b) = g(c,d) \implies (a,b) = (c,d)$,  then $g$ is an injective function.   $g(a,b) = (\frac{a^2 + b^2}{a^2 + b^2 + 1}a, \frac{a^2 + b^2}{a^2 + b^2 + 1}b)$, $g(c,d) = (\frac{c^2 + d^2}{c^2 + d^2 + 1}c, \frac{c^2 + d^2}{c^2 + d^2 + 1}d)$  Since $g(a,b) = g(c,d)$, $\frac{a^2 + b^2}{a^2 + b^2 + 1}a = \frac{c^2 + d^2}{c^2 + d^2 + 1}c$.  Furthermore, $0 \leq \frac{a^2 + b^2}{a^2 + b^2 + 1} &lt; 1$ and $0 \leq \frac{c^2 + d^2}{c^2 + d^2 + 1} &lt; 1$.  I want to have $\frac{a^2 + b^2}{a^2 + b^2 + 1} = \frac{c^2 + d^2}{c^2 + d^2 + 1}$,  and for that I need  $\frac{a^2 + b^2}{a^2 + b^2 + 1} \leq \frac{c^2 + d^2}{c^2 + d^2 + 1}$ and $\frac{c^2 + d^2}{c^2 + d^2 + 1} \leq \frac{a^2 + b^2}{a^2 + b^2 + 1}$.  which I am not so sure how to get.  Is this a valid approach to reaching the conclusion $a = c$?  Thank you for reading.
120744
Consider the subset $B=\{(x,y):x^2+y^2\leq 1\}\ \subseteq \mathbb{R}^2$. Show that $\vert B \vert = \vert \mathbb{R}^2 \vert$.  It is easy to see tht $f:B\rightarrow\mathbb{R}^2$ defined as $f(x,y)=(x,y)$ is injective.   However, in the book I'm working out of (Book of Proof by Richard Hammack), the other injection noted is  $g:\mathbb{R}^2\rightarrow B$ defined as  $$g(x,y)=\left(\frac{x^2+y^2}{x^2+y^2+1}x,\frac{x^2+y^2}{x^2+y^2+1}y\right).$$  In the book I'm asked to verify this is indeed an injective function.  My approach was to assume $g(x,y)=g(m,n)$ and through basic algebra I was able to deduce  $x(m^2+n^2+1)(x^2+y^2)=m(x^2+y^2+1)(m^2+n^2)$  and  $y(m^2+n^2+1)(x^2+y^2)=n(x^2+y^2+1)(m^2+n^2)$  but how does this show $(x,y)=(m,n)$? Or is this the wrong approach?
17805
Let $f(x)$ and $g(x)$ be two continuous functions on $[0,1]$ and  $$\int_{0}^{1}f(x) dx= \int_{0}^{1}g(x)dx = 1$$   Show that there exist  $[a,b]\subset [0,1]$, such that  $$\int_{a}^{b}f(x) dx= \int_{a}^{b}g(x)dx = \frac{1}{2} $$  The question can be solved by considering the fundamental group of $S^1$, now I am wondering if we can solve it by real-analysis.  Here is the topological solution:  Assume that for any $[a,b]\subset[0,1]$, we always have  $$(\int_{a}^{b}f(x) dx\neq \frac{1}{2}) \quad \vee \quad(\int_{a}^{b}g(x)dx \neq \frac{1}{2}) $$ Consider mapping $$\phi:D\rightarrow\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\},\,\,(x,y)\mapsto(\int_{y}^{x}f(t) dt,\int_{y}^{x}g(t) dt)$$ where $D=\{(x,y)|0\leq x\leq y\leq 1\}$.  Let $a$ be path from $(0,0)$ to $(0,1) $in $D$ and $b$ be path from $(0,1)$ to $(1,1) $in $D$, then $ab$ is a path from $(0,0)$ to $(1,1) $in $D$ and $$\phi\circ (ab):[0,1]\rightarrow\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\} $$ Notice that $$(\phi\circ (ab))(0)=(\phi\circ (ab))(1)=(0,0)$$  so $\phi\circ (ab)$ is a loop based on $(0,0)$ in $\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\}$.  For any $t\in [0,\frac{1}{2}]$, it's not hard to get that $$(\phi\circ (ab))(t+\frac{1}{2})=(1,1)-(\phi\circ (ab))(t)$$ is equivalent to $$(\phi\circ (ab))(t+\frac{1}{2})-(\frac{1}{2},\frac{1}{2})=-((\phi\circ (ab))(t)-(\frac{1}{2},\frac{1}{2}))$$  Define retraction  $$r:\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\}\rightarrow S^1,\quad (x,y)\rightarrow\ \frac{(x,y)-(\frac{1}{2},\frac{1}{2})}{||(x,y)-(\frac{1}{2},\frac{1}{2})||}$$  Then $r\circ \phi\circ (ab)$ is a loop on $S^1$, such that $$(r\circ \phi\circ (ab))(t+\frac{1}{2})=-(r\circ \phi\circ (ab))(t),\quad\forall t\in [0,\frac{1}{2}]$$  So $&lt;r\circ \phi\circ (ab)&gt;$ is not trivial in $\pi_1(S^1)$.   However, $ab$ is path homotopic to $c$ in $D$, where $c$ is the path between $(0,0)$ and $(1,1)$ in $D$. In this case, $r\circ \phi\circ (ab)$ is a point-path in $S^1$ by $$(\phi\circ c)(t)=\phi(t,t)\equiv (0,0)$$  which leads to contradiction.
115396
I have a problem with the next exercise. It's so confusing for me. I have one implication, but no the other.   Let $f:D\subset\mathbb{R}^2\rightarrow\mathbb{R}$ with $D$ open set. The exercise is the next:  $\displaystyle\frac{\partial f}{\partial x} = \displaystyle\frac{\partial f}{\partial y}$ for all $(x,y)\in D$ if and only if there exist a derivable function $g$ such that $f(x,y)=g(x+y)$ for all $(x,y)\in D$.  $\Leftarrow )$   Let $(x,y)\in D$  $\displaystyle\frac{\partial f}{\partial x}=\lim\limits_{h\rightarrow 0} \displaystyle\frac{f(x+h,y)-f(x,y)}{h}=\lim\limits_{h\rightarrow 0}\displaystyle\frac{g(x+y+h)-g(x+y)}{h}=g'(x+y)$ (the limit there exist because $g$ is derivable).  $\displaystyle\frac{\partial f}{\partial y}=\lim\limits_{h\rightarrow 0} \displaystyle\frac{f(x,y+h)-f(x,y)}{h}=\lim\limits_{h\rightarrow 0}\displaystyle\frac{g(x+y+h)-g(x+y)}{h}=g'(x+y)$ (again, the limit there exist because $g$ is derivable).  Thus, $\displaystyle\frac{\partial f}{\partial x} = \displaystyle\frac{\partial f}{\partial y}$  But, I can't understand the implication $\Rightarrow) $. Some hint? I think define a function using the partial derivatives, but, really, I don't know how.
23743
Let $f,g:$ $R \rightarrow R$ where $g(x)$ $=$ $1$ $-$ $x$ $+$ $x^2$ and $f(x)$ $=$ $ax$ $+$ $b$  If $g(f(x))$ $=$ $9x^2$ $-$ $9x$ $+$ $3$, determine $a$ and $b$.  So far, I have "fit" $f(x)$ into $g(x)$ as follows: $g(f(x))$ $=$ $(ax + b)^2$ $-$ $(ax+b)$ $+$ $1$  Is there a relationship between compositions and inverse functions I should be using to isolate $a$ and $b$? How does one extract those variables in order to determine their value?  Thank you!
15420
Let $f,g:[a,b]\subset \mathbb{R} \to \mathbb{R} $ be continuous functions with $f^2(x)=g^2(x)\neq 0$ for all $x\in[a,b]$. Show $f=g$ or $f=-g$ on $[a,b]$.
108117
Find all functions $g:\mathbb{R}\to\mathbb{R}$ with $g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$.  I think the solutions are $0, 2, x$. If $g(x)$ is not identically $2$, then $g(0)=0$. I'm trying to show if $g$ is not constant, then $g(1)=1$. I have $g(x+1)=(2-g(1))g(x)+g(1)$. So if $g(1)=1$, we can show inductively that $g(n)=n$ for integer $n$. Maybe then extend to rationals and reals.
48713
Let $f:\Bbb{R}\to\Bbb{R}$ be a function given by  $$f(x)=\begin{cases} \exp\left(\frac{1}{x^2-1}\right) &amp; \text{if }\vert x\vert\lt 1\\  0 &amp; \text{if }\vert x\vert\geqslant 1  \end{cases}$$  I would like to prove that $f\in C^\infty$, that is, $f\in C^k$ for all $k\in \mathbb{N}$. I think that it can be done by induction on $k$. If $\vert x\vert\gt1$, the problem is trivial. On other points, the base case is the simplest and the only that I'm be able to do. Can someone help me?  Thanks.
81180
I am trying to prove the following statement but have trouble  comprehending/going forward with some parts! Here is the statement:     If $A$ is any set, then  $|A|$ $&lt;$ $|P(A)|$   Here is what I have so far:  We need to show that there is an injection from $A$ to $P(A)$ but not a surjection.   A natural choice for an injection is the function $ f(x)$ $=$ $\{x \}$, which in plain English, takes any element $x$ (that is in $A$) and sends it to the one-element set $\{x \}$. Thus $f(x)$ is injective!  To show that there is no surjection, for the sake of contradiction, assume there is a surjection. Here is where I start to have trouble. Surjectivity means that every element of the co-domain is mapped to an element of the domain, correct? Consequently, in this particular case, we are "matching" sets (from $P(A)$) to elements (from $A$) right?  If the above is correct, my problem arises here. I am not sure how to prove that $f$ is not surjective. Unfortunately, I am easily confused by notation so please explain in English.  Thank you in advance!! :)
13730
Define a function $f\colon\mathbb{R}\to\mathbb{R}$ which satisfies $$f(xy)=f(x)f(y)-f(x+y)+1$$ for all $x,y\in\mathbb Q$. With a supp condition $f(1)=2$. (I didn't notice that.)  How to show that $f(x)=x+1$ for all $x$ that belong to $\mathbb{Q}$?
6199
For any pair of sets $A$ and $B$, we can define $A\le B$ iff there exists injection $f\colon A\rightarrow B$. I am trying prove that  $$(A\le B)\vee (B\le A).$$  I have tried assuming  $\neg (A\le B)$, then proving $B\le A$ by constructing the required injection, but I haven't been able to make any progress. Any hints, etc. would be appreciated.  EDIT  Assuming  $\neg (A\le B)$, can you prove there exists a surjection $f: A\rightarrow B$? Then it would be easy, by applying AC, to construct an injection $g: B\rightarrow A$
101779
Let $f \colon M \to N$ a function. Show that the following affirmations are equivalent: (a) $f$ is injective (b) every two subsets $M_1$ and $M_2$ of $M$, $f(M_1 \cap M_2)=f(M_1) \cap f(M2)$  (a) $\implies$ (b): Let $y \in f(M_1 \cap M_2)$ $\implies$ exists $x$ an element from $M_1 \cap M_2$ such that $f(x)=y$ $\implies$ exists $x \in M_1$ such that $f(x)=y$ and exists $x \in M_2$ such that $f(x)=y$ $\implies y \in f(M1) \cap f(M2)$. So, $f(M_1 \cap M_2)$ is a subset of $f(M_1) \cap f(M_2)$. Let $y \in f(M_1) \cap f(M_2)$. How can I prove, using the injectivity of the function $f$,that $y \in f(M_1 \cap M_2)$?  (b)$\implies$ (a) $f(M_1 \cap M_2)=f(M_1) \cap f(M_2)$ means that every $y$ from $f(M_1 \cap M_2)$ belongs to $f(M_1) \cap f(M_2)$ and  every $y$ from $f(M_1) \cap  f(M_2)$ belongs to $f(M_1 \cap M_2)$. How can I prove that $f$ is injective?
38630
I've already done a lot of searching (in particular: ) to try to prove this statement:      $f: A \to B$ is injective if and only if it has a left inverse. [I'm going to also assume $A$ and $B$ are nonempty.]   My attempt so far:     $\Leftarrow$ Suppose $f$ has a left inverse and $f(a_1) = f(a_2)$ for    $a_1, a_2 \in A$. Then there is a function $g: B \to A$ such that   $g\left(f(a_1)\right) = a_1$ and $g\left(f(a_2)\right) = a_2$. Since   $g$ is well-defined, it follows that $g\left(f(a_1)\right) =  g\left(f(a_2)\right)$, and hence $a_1 = a_2$.         $\Rightarrow$Now   suppose $f$ is injective. Then $f(a_1) = f(a_2) \implies a_1 = a_2$.   Furthermore, since $A$ is nonempty, there is an element $x \in A$.   Define $g: B \to A$ by  \begin{equation*} g(b) = \begin{cases}  f^{-1}(b), &amp; b \in \text{Im}(f) \\ x, &amp; b \in A \setminus  \text{Im}(f)\text{.} \end{cases} \end{equation*} This mapping is   well-defined, since if $f$ is injective, $f^{-1}(b)$ is unique.   My main question: does this imply that if $f: A \to B$ is injective that any mapping $g: \text{Im}(f) \to A$ is bijective? Otherwise, I don't understand why we can even use $f^{-1}$ on an element. [Please read the link above for more details - in proof 1.]
30480
For $f \in C[a,b], g\ge0$ (or $g\le0$) on $[a,b]$, then there exists $c\in(a,b)$ $\int _a^b f(x)g(x)dx = f(c)\int_a^bg(x)dx$.   It is MVT for integral.  Why $c \in (a,b)$ ? How about $c\in[a,b]$? Is it wrong?  I am wondering this because, in proving progress,   by IVT, there exist $m,M$ s.t $m\le f(x)\le M$   so $m\int_a^bg(x)dx$ $\le$ $\int_a^b f(x)g(x)dx$ $\le$ $M\int_a^bg(x)dx$.  when $\int_a^bg(x)dx&gt;0$ then $m\le$ $\frac{\int_a^b f(x)g(x)dx} {\int_a^bg(x)dx}$   $\le$  $M$ and   $\frac{\int_a^b f(x)g(x)dx} {\int_a^bg(x)dx}$ becomes $f(c)$, so it seems $c\in [a,b]$ might work.  So, I want to know why $c\in(a,b)$, or $c\in[a,b] $ would be okay.
40530
Consider $f:A \to B$ and $g: B \to C$. If I know that both $f$ and $g$ are injective, then can I state that $$g  \circ f:A \to C$$ is injective?
78422
From wikipedia I obtain the following definition of an injective function :     Let $f$ be a function whose domain is a set $A$. The function $f$ is injective if for all $a$ and $b$ in $A$, if $f(a) = f(b)$, then $a = b$; that is, $f(a) = f(b)$ implies $a = b$.    From this I conclude that a function $f$ is injective if the below statement is true for all $a,b \in A$:  $$f(a)=f(b) \implies a=b$$  My question is: Can I re-formulate the above statement as $f(a)=f(b) \iff a=b$ ?
109549
How can I proof this:     Show that no application $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, of $C^k$ class, $k \geq 1$ can be injective, i.e., there are $A,B \in \mathbb{R}^2$ such that $A \neq B$ and $f(A) = f(B)$ (Hint: show that given $(a,b) \in \mathbb{R}^2$ in every open neighbourhood of $(a,b)$ there are points $(c,d),(c_1,d_1)$ where ${\partial}_1f(c,d) \neq 0$ or ${\partial}_1f(c_1,d_1) \neq 0$).   The problem is that I have no idea how to use the hint.  Thanks for all your comments.
6536
The functions of each pair are inverse to each other. For each pair, check that both compositions give the identity function.  $F: \mathbb{R} \to \mathbb{R}$ and $F^{−1}:\mathbb{R} \to \mathbb{R}$ are defined by $F(x)=3x+2$ and $F^{−1}(y)=\dfrac{y−2}{3}$. for all y ∈ R  My attempt:     Inverse Function  For each particular but arbitrarily chosen $y \in \mathbb{R}$, according to the definition of $f^{-1}$, $f^{-1}(y) = \dfrac{y-2}{3}$ is a unique real number $x$ such that $f(x) = y$. \begin{align*} F(x) &amp; = y\\ 3x + 2 &amp; = y\\ x &amp; = \frac{y-2}{3} \end{align*} Therefore, $f^{-1}(y) = \frac{y-2}{3}$.  Compositions of Functions  The functions $g \circ f$ and $f \circ g$ are defined as follows: $$(g \circ f)(x) = g(f(x)) = g(3x + 2) = 3x + 2$$  for all $x \in \mathbb{Z}$.