Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be an injective and continous function. Prove that $f$ is monotone.  My proof-trying Since $f$ is injective for all $x_{1},x_{2}\in\mathbb{R}$ there is a $y$ in $\mathbb{R}$ such that $f\left( x_{1}\right) =f\left( x_{2}\right)$. Since $f$ is continous for any $\varepsilon &gt;0$ there is $\delta &gt;0$ such that if $\left| x-a\right| &lt; \delta$ then we have $\left| f\left( x\right) -f\left( a\right) \right| &lt; \varepsilon$.  So how can I show $f$ is monotone?
14021
I would like to prove the statement in the title.  Proof: We prove that if $f$ is not strictly decreasing, then it must be strictly increasing. So suppose $x &lt; y$.  And that's pretty much how far I got.  Help will be appreciated.
74204
Let $f:\mathbb R \to \mathbb R$ be a continuous function such that $|f(x)-f(y)|≥|x-y| ,\forall x,y \in \mathbb R  $ , then how do we prove that $f$ is surjective ?
42395
If $f$ is continuous from $ [0,1]$ to itself and $f(0)=0$, $f(1)=1$ and $f(f(x))=x$ then show that $f(x)=x$ for every $ x $ in the domain of function.  I'd like to prove this statement, my idea goes like below:  Suppose that the converse of the statement holds. then without loss of generality, assume that there exists some $c$ in $[0,1]$ such that $d=f(c)&gt;c$. By symmetry, $f(d)=f(f(c))&lt;f(c)=d$. And $f$ is bijective because $f$ has an inverse function in this domain. So $f$ should be strictly increasing function. But on $[c,d]$, $f$ is decreasing. This is a contradiction.  So it should be that $f(x)=x$ on $[0,1]$.  Is this argument right? And I'd like to know another argument which prove this statement. Thank you for your answer in advance :)
42554
Find all the continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that if $x-y\in \mathbb{Q}$ then $f(x)-f(y)\in \mathbb{Q}$.   Can somebody give me some hint to solve this problem. I think that the functions are $f(x)=mx+b$ where $m\in\mathbb{Q}$.  Thanks!
101469
Suppose that $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous on $\mathbb{R}$ and that $f(r)=0$ for all $r \in \mathbb{Q}$. Prove that $f(x)=0$ for all $x \in \mathbb{R}$.   My attempt: Define a sequence $(x_n)$ where $x_n \in \mathbb{Q}$ for all $n \in \mathbb{N}$ and assume that $(x_n) \rightarrow a \not\in \mathbb{Q}$. Since $f$ is continuous, we have $\lim_n{f(x_n)}=f(a)=0$. Since $a$ is arbitrary irrational number, we have $f(a)=0$ for all $a \not\in \mathbb{Q}$. Hence, we proved the statement.  Is my proof valid? or is there any flaw ?
46046
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function s.t $|f(x)-f(y)|\ge \frac{1}{2}|x-y|$ for all $x,y$. Then $f$ is injective can be seen easily as $x\neq y$ implies $f(x)\neq f(y)$.  But how to check that $f$ is surjective or not ?
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If $f:\mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(x)=x$ has no real solution, then show that $f(f(x))=x$ has no real solution either.   Is the proof trivial as it seems or does it need an analytical approach?
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.]
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I'm having trouble with this functional equation, to find all functions $f:\mathbb R \to \mathbb R$ for which the following is valid for all $x,y\in \mathbb R$: $f(x) + f(x+f(y)) = y + f(f(x) + f(f(y)))$ I haven't been able to pinpoint any solution, nor to prove surjectivity or injectivity. So 0 progress. Tried with $x=y=0, x=0, y=0, x=-f(y), y=f(x)$, and even with some assumptions on bijectivity, but nothing helped. I don't know the source but seems like a pretty difficult problem, maybe even impossible to solve.
94294
Firstly, the assignment:     Let $a,b \in\mathbb{R}$ and $a &lt; b$. Furthermore let $f: [a,b] \rightarrow [a,b]$ be monotone increasing. Show that if $x:= \mathbf {sup}\{y \in [a,b] \| \ y ≤ f(y)\}$ then $f(x) = x$.   It's good on me that the assignment itself actually tells you what $x$ is so I can prove the theorem by showing that for $\forall x' \not= x: f(x) \not= x$ and I'm done. However, I have not arrived at a contradiction so far.  Any tips?
65877
Let $f: \mathbb R\to \mathbb R$ be a continuous function such that $f$ attains neither its minimum nor its maximum at any open interval $I \subseteq  \mathbb R$ , then how to prove that $f$ is monotone ?
101313
Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function. $x\in\mathbb{R}$ is a fixed point of $f$ if $f(x)=x$. Show that if $f'(t)\neq 1\;\forall\;t\in\mathbb{R}$, then $f$ has at most one fixed point.  My biggest problem  with this is that it doesn't seem to be true. For example, consider $f(x)=x^2$. Then certainly $f(0)=0$ and $f(1)=1 \Rightarrow 0$ and $1$ are fixed points. But $f'(x)=2x\neq 1 \;\forall\;x\in\mathbb{R}$. Is there some sort of formulation that makes this statement correct? Am I missing something obvious? This is a problem from an old exam, so I'm assuming that maybe there's some sort of typo or missing condition.
18925
I'm studying measure theory and two question came to my mind:        If $f:\mathbb{R}\to\mathbb{R}$ is monotone and $B\subseteq\mathbb{R}$ is borel, is the image $f(B)$ borel?   If $f:\mathbb{R}\to\mathbb{R}$ is a monotone function (say, non-decreasing), does there exist a sequence of continuous functions $f_n:\mathbb{R}\to\mathbb{R}$ converging pointwise to $f$?      Here's the motivation for those questions: Let $(X,M)$ and $(Y,N)$ be measure spaces.   It's known that if $\mu$ is a measure on $(X,M)$ and $f:X\to Y$ is measure, then we have the pushforward measure $f_*\mu(A)=\mu(f^{-1}(A))$ on $(Y,N)$. What if we were to define a "pullback measure"? Given a function $f:X\to Y$ such that $f(M)\subseteq N$ (i.e. $f$ maps measurable sets to measurable sets) and a measure $\nu$ on $N$, the natural formula would be $f^*\nu(A)=\nu(f(A))$. So we ask if is there a good amount of functions which map measurable sets to measurable sets in $\mathbb{R}$, and monotone functions seem like good candidates (for strictly monotone, continuous functions, the result is valid and there are several answers on the web). If this were true, maybe we could use some convergence argument to solve the problem above.
38375
Prove that $[0,1]$ isn't homeomorphic to $\mathbb{R}$  My first thought is that there can not be a continuous bijection, $f$, from $[0,1]$ to $\mathbb{R}$ because a continuous function that maps $[0,1]\rightarrow \mathbb{R}$ must be bounded so there can't be a surjection. So that would then be the proof.   Though I believe I am incorrect in my thinking because a hint for the assignment says to use the intermediate value theorem. That is "Suppose $f:[a,b]\rightarrow\mathbb{R}$ is continuous. If $f(a)&lt;\delta&lt;f(b)$ or $f(b)&lt;\delta&lt;f(a)$ then $\delta=f(c)$ for some $c\in[a,b]$".   Why is my first thought wrong and how is the IVT useful?
53589
If $f$ is continous in $[a,b] \subset \mathbb{R}$ and exist real numbers $\alpha \not= \beta$ such that $$\alpha\int_{a}^{c} f(x) dx + \beta\int_{c}^{b} f(x) dx  =0$$ for all $c \in [a,b]$ show that $f (x)=0$ for all $x \in [a,b].$
44719
Assume $f$ is a continuous one-to-one function over an interval. Prove that $f$ is strictly monotone.   Attempt  Since we know that $f$ is one-to-one, for every $f(x)$ there is exactly one element $x_0$ that maps to it. Thus, if $f(x) = f(y)$, then $x=y$. Now let's suppose that $f$ isn't monotone. That is, it goes from nonincreasing to nondecreasing. Thus, on some interval $[a,b]$ we must have  $f'(x) &lt; 0$ and on some other interval $[c,d]$ we must have $f'(x) &gt; 0$. As a result we must have that $f'(x) = 0$ at some point. How do I show this contradicts the definition of $f$?
111680
Suppose $f'(x)=0$ for all $x\in (a,b)$. Prove that $f$ is constant on $(a,b)$.   This seems painfully obvious, but I can't prove it rigorously.  $f'(x)=0$ for all $x\in (a,b)$ means that for any $c\in(a,b)$, we have $$\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}=0$$ So for any $c\in (a,b)$, for any $\epsilon$ there exists $\delta$ such that if $|x-c|&lt;\delta$ then $$\left|\frac{f(x)-f(c)}{x-c}\right|&lt;\epsilon$$ which also means $$|f(x)-f(c)|&lt;\epsilon\delta$$ How to continue from here?