I am stuck at this problem for long time:    Prove that if $f:\mathbb{R}\to\mathbb{R}$ is some real-valued function, Then $|f(\mathbb{N})|\leq\aleph_0$.  In other words, prove that there exists a one-to-one function from $f(\mathbb{N})=\{f(n)|n\in\mathbb{N}\}$ to $\mathbb{N}$.    If $f(\mathbb{N})$ is finite then it is clear that $|f(\mathbb{N})|\leq\aleph_0$, But if $f(\mathbb{N})$ is infinite then I got stuck.  Thanks for any hint/help.
122676
Hi everyone I'd like to know if the following reasoning is correct, any suggestion would be great. Thanks.  Proposition: Let $Y$ be a set and let $f: \mathbb{N}\rightarrow Y$ be a function. Then $f[\mathbb{N}]$ is at most countable.   Proof: If $Y$ is finite the result follows, this is because$f[\mathbb{N}]\subset Y$ and any subset of a finite set  is finite and has at most the size of the set in which is contained.   We may assume  that the set $Y$ is infinite. We set $A:= \{\,n\in\mathbb{N}:f(i)\not=f(n) \,\,\text{ for  any }i&lt; n\, \}$. And we define the function $g:A \rightarrow f[\mathbb{N}]$ as the restriction of $f$ in $A$. We claim that the function $g$ is a bijective map. From the definition of $A$ we already know that the map is $1:1$. We will show that also is onto.   Let $y\in f[\mathbb{N}]$. Suppose for the sake of contradiction that there is no $n\in A$ such that $g(n)=y$. Since $y$ lies in the image of $f$, then there is some $j\in \mathbb{N}$ for which $y=f(j)$. Then either $f(j)\not= f(i)$ for any $i&lt;j$ or there exists some $i&lt;j$ such that $f(j)= f(i)$. For the former, $j$ lies in $A$ (by its definition). For the latter, the set $\{\, n\in \mathbb{N}: f(j)=f(n)\, \text{and }\,n\not=j \}$ is non empty, and hence has a minimum element. Let $m$ be its minimum element. Then $f(m)\not=f(n)$ for any $n&lt;m$ and thus $\,m\in A$ but $g(m)=f(m)= f(j)=y$. Thus any case leads to a contradiction. The result follows by reductio ad absurdum. Hence $f$ is surjective and injective, i.e., a bijective map.   Since there is a bijection between $A$ and the images of the natural numbers under $f$, both have the same cardinality. Also we already know that $A\subset \mathbb{N}$ and any set of the natural numbers is denumerable, thus $f[\mathbb{N}]$ is countable. $\Box$
111761
Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be an holomorphic function such that $f'(z_0) \neq 0$ for some $z_0 \in \mathbb{C}$.Prove that there is $r&gt;0$ such that, if $|z-z_0|&lt;r$ and $z \neq z_0 $, then $f(z) \neq f(z_0)$.  It seems very trivial, but I can't solve it...
123563
If $f:\mathbb{R}\to\mathbb{R}$ is continuous and $$\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=\infty$$  Prove that $f$ has a minimum on $\mathbb{R}$; that is, there is an $x_m\in\mathbb{R}$ such that $$f(x_m)=\inf_{x\in\mathbb{R}}f(x)&lt;\infty$$  I think the best approach would be to prove the existence of a lower bound, that is $\exists M\in\mathbb{R}:M&lt;f(x),\forall x\in\mathbb{R}$, but I don't know how to do this. I am probably supposed to use the extreme value theorem or the intermediate value theorem, but I don't know how to do that.
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Do there exist functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y),$ but which aren't linear? I bet you they exist, but I can't think of any examples.  Furthermore, what hypotheses do we need to put on $f$ before no such functions exist? I feel continuity should be enough.
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Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ having the following properties?   $f(x) = 0$ for all $x \le 0$. $f(x) = 1$ for all $x \ge 1$. For $0 &lt; x &lt; 1$, $f$ is strictly increasing. $f$ is everywhere $C^\infty$. The sequence of $L^\infty$ norms $\langle \left\lVert f \right\rVert_\infty, \left\lVert f' \right\rVert_\infty, \left\lVert f'' \right\rVert_\infty, \dots \rangle$ is bounded.   If we impose only the first four conditions, there is : for $0 &lt; x &lt; 1$, define $f(x)$ by $$ f(x) = \frac{e^{-1/x}}{e^{-1/x} + e^{-1/(1-x)}} = \frac{1}{1 + e^{1/x - 1/(1-x)}} $$ However, the derivatives of this function appear to grow quite rapidly. (I'm not sure how to verify this, but it seems at least exponential to me.)  If such a function does not exist, what is the smallest order of asymptotic growth that the sequence $\langle \left\lVert f \right\rVert_\infty, \left\lVert f' \right\rVert_\infty, \left\lVert f'' \right\rVert_\infty, \dots \rangle$ can have?
51298
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$(x+y)(f(x)-f(y))=(x-y)f(x+y)$$  My attempt:   If $x=-y \not = 0$ then $0= 2x f(0)$ so $f(0)=0$.   Suppose for the sake of contradiction that $f(x)=f(x+\epsilon)$ for some $x$ and $\epsilon&gt;0$. Let $y=x+\epsilon$. Then $$0=\epsilon \cdot f(2x+\epsilon)$$ therefore as $2x+ \epsilon$ can take any real value $f$ is either strictly increasing or strictly decreasing or $f(x)=0 \; \; \forall \; \;x \in \mathbb{R}$.   Note that $f(x)=ax$ is a solution $\forall \; \;a \in \mathbb{R}$. Thanks so much for any help!
45963
1) Construct a continuous function $f$ on $\mathbb{R}$ that is integrable on $\mathbb{R}$ but $\displaystyle\limsup_{x \to \infty} f(x)  = \infty$.  I took the function that is equal to $n$ on $[n, n + 1/n^{3})$ and made it continuous by saying that $f$ is the line segment joining $n$ and $n+1$ on $[n + 1/n^{3}, n+1)$. But I am failing to prove this integrable. For this, $\lim f(x) = \infty$ but how do you prove in general, if limit does not exist that $\limsup$ is infinity?  2) Prove that if $f$ is uniformly continuous and integrable on $\mathbb{R}$ we have $\displaystyle\lim_{|x| \to \infty} f(x) = 0$.  Any help is appreciated.  Thanks
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I need help solving the next question, I think that it's actually a refutation:     let $f,g: \mathbb{N} \rightarrow \mathbb{N}$  so that $\forall n ∈ \mathbb{N} : f(n) = g(2n)$   Prove that if $f$ is surjective $g$ is not injective
115504
Problem: Suppose that $f$ is a bounded, real-valued function on $[a,b]$ such that $f^2\in R$ (i.e. it is Riemann-Integrable). Must it be the case that $f\in R$ ?      Thoughts: I think that this is not necessarily true, but I am having trouble refuting or even proving the above. Of course, the simplest way to prove that it is not necessarily true would be to give an example, but I am unable to think of one! I also have tried using $\phi(y)=\sqrt y$ and composing this with $f^2$ (to try show $f$ is continuous); however, the interval $[a,b]$ may contain negative numbers so I can't utilise $\phi$ in that case.  Question: Does there exist a function $f$ such that $f^2\in R$ but $f$ $\not\in R$ ? Or conversely, if $f^2\in R$ does this always imply $f$ $\in R$ ? (If so, could you provide a way of proving this).
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Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ that takes each value in $\mathbb{R}$ three times? If not, how could I prove that such a function does not exist?
94508
Some very wise people here have just told me that  $\aleph_0 = \mathbb{N}$, i.e. that the cardinality of the set of natural numbers is just the set of natural numbers itself. Is this now the general consensus in mathematics, real analysis in particular? Or have I got completely the wrong end of the stick as usual?
28012
I am working on this exercise for an introductory Real Analysis course:     Show that |$\mathbb{R}$| = |$\mathbb{R}^2$|.   I know that $\mathbb{R}$ is uncountable. I also know that two sets $A$ and $B$ have the same cardinality if there is a bijection from $A$ onto $B$. So if I show that there exists a bijection from $\mathbb{R}$ onto $\mathbb{R}^2$ then I beleive that shows that |$\mathbb{R}$| = |$\mathbb{R}^2$|.   Let $x_i \in \mathbb{R}$, where each $x_i$ is expressed as an infinite decimal, written as $x_i = x_{i0}.x_{i1}x_{i2}x_{i3}...,$. Each $x_{i0}$ is an integer, and $x_{ik} \in \left \{ 0,1,2, 3, 4, 5, 6, 7, 8, 9 \right \}$. Then, let  $$f(x_i)=(x_{i0}.x_{i1}x_{i3}x_{i5}... ,x_{i0}.x_{i2}x_{i4}x_{i6}...)$$  What should I do to show that $f: \mathbb{R} \to \mathbb{R}^2$ is an injective function? Any suggestions or help with the question would be appreciated.
45536
Let $f \in \mathbf R[x]$  and for every $k \in \mathbf Z$, $f(k)\in \mathbf Z$. Does this imply that $f \in \mathbf Z[x]$?   I tried to find a counter example, but my intuition tells me it's true and I don't know how to prove the statement either.
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Question     Let the function $f:\mathbb R\to\mathbb R$,and such   $$f(xf(y)+x)=xy+f(x)$$      Find all $f(x)$   Let $x=1,y=1$,then $$f(f(1)+1)=1+f(1)$$ let $f(1)=t$,then $$f(t+1)=1+t$$  So I guess $$f(x)=x$$ is such it, and I found $f(x)=-x$ is also such it?   Can you help?  I found this simaler problem：(BMO 1997, 2000) $$f(xf(x)+f(y))=y+(f(x))^2,f:R\to R$$ this answer $f(x)=x$ or $f(x)=-x$,  the full solution can see
114828
Let $f: \mathbb{R} \to D \subseteq \mathbb{R}$ be a continuous function. Is there a function $f$ that satisfies the following property? $\forall y \in D$, there are exactly 3 $ x_1,x_2,x_3 \in \mathbb{R}$ such that $f(x_1) = f(x_2) = f(x_3) = y$.
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Let $f : \mathbb{R} \to \mathbb{R}$ be a function which is differentiable everywhere on $\mathbb{R}$. If we know that  $$\lim_{x\to-\infty}f(x) = L = \lim_{x\to\infty}f(x)$$ How would one go about showing that there must exist $c \in \mathbb{R}$, such that $f'(c) = 0$ for all cases at once?
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I'm annoyed by quite a simple problem in calculus (I apologize in advance if I'm not using adequate terms in English, I don't take the course in English nor am I a native speaker):  Let $f:\mathbb R\longrightarrow \mathbb R$ be a function which is continuous at $0$ and satisfies $f(x)=f(2x)\,\forall x\in\mathbb R$.  My question is how do I formally prove that for every $x&lt;0$ $f(x)=c_1$, and for every $x&gt;0$ $f(x)=c_2$, where $c_1,c_2\in \mathbb R$? After proving those two statements (or at least one of them WLOG) it would be quite easy to prove that $f(0)=c_1=c_2$ by the definition of continuity of $f(0)$ and thus $f(x)$ is constant. Those two statements are just so obvious to see that I can't think of any formal way of proving so without "cheating". Thanks in advance!