I have the following function defined on $\mathbb{R}$:  $$f(x) = \begin{cases}     0 &amp; \text{if $x$ irrational} \\     1/n &amp; \text{if $x = m/n$ where $m, n$ coprime}   \end{cases}$$  I want to show that $f$ is continuous at every irrational point, and has a simple discontinuity at every rational point. I was able to show the first and partially the second (I showed that $f$ has a discontinuity at every rational point, but got stuck on showing that the discontinuity is simple).  However, I've realized that I can simply show that $\lim_{t \rightarrow x}f(t) = 0$ for every $x$, and both of the things I want to show follow from this. How can I show this?
94393
I have to prove that a function $f:]0,1] \rightarrow \Bbb R$ : $$ f(x) = \begin{cases} \frac1q,  &amp; \text{if $x \in \Bbb Q$  with $ x=\frac{p}q$ for $p,q \in \Bbb N$ coprime} \\ 0, &amp; \text{if $x \notin \Bbb Q $} \end{cases} $$  is discontinuous in every point $x \in \ ]0,1] \cap\Bbb Q$.  And then to consider $x \in \ ]0,1] \backslash \Bbb Q$ and prove that it is continuous.  For now I learned different ways to prove continuity (epsilon-delta, sequences), but I'm never sure what would be better to use in each different case.  I wanted to prove the discontinuity by using sequences: $$\forall x_n \quad x_n\rightarrow a \quad \Rightarrow \quad f(x_n) \rightarrow f(a)$$  I tried creating a sequence $ x_n=\frac1n + a$, we know it converges to $a$ but $f(x_n)\ $ doesn't  converges to $\ f(a)$ because there would still be some points not in our set (irrational numbers that creates gaps).  But I don't think it works, so I'm asking you if you could help me solving the two questions.
87288
The question is: Prove that if $f:\mathbb{R}\to\mathbb{R}$ is continuous and $f(x) = 0$ whenever $x$ is rational, then $f(x) = 0$ for all $x\in\mathbb{R}$.  My proof:  Let $ x\in\mathbb{R} $. If $ x\in\mathbb{Q}$, then $ f(x)=0 $.  Otherwise $ x\in\mathbb{R\backslash Q} $. Then there exists a sequence of rational numbers $ (q_n)_{n\in\mathbb{N}}$ (for example $q_n= \frac{\lfloor{nx\rfloor}}{n} ,\forall n\in\mathbb{N}$) such that $q_n\to x$ as $n\to \infty$. By the properties of $f$, we have $f(q_n)=0$ for all $n\in\mathbb{N}$. So $f(q_n)\to 0$ as $n\to \infty$. Since $f$ is continuous, $q_n\to x$ implies that $f(q_n)\to f(x)=0$. So $f(x) = 0$ for all $x\in\mathbb{R}$.   Thank you for any help.
33321
The following is not a homework problem. I am doing it for self study.      Prove that any continuous function from $\mathbb{R}$ to $\mathbb{Q}$ is constant.    Here is my proof:     Let $f:\mathbb{R}\rightarrow \mathbb{Q}$ be such a function.    ? We first show that $\mathbb{Q}$ is disconnected. Let $p$ be an irrational number. Then we can write $\mathbb{Q}$ as  $$ \big( \mathbb{Q} \cap (-\infty,p) \big) \cup \big( \mathbb{Q} \cap (p,\infty) \big),  $$  which is a disconnection of $\mathbb{Q}$.      Moreover, $\mathbb{Q}$ is also totally disconnected, as any open subset $U$ of $\mathbb{Q}$ must contain two rational numbers and there is always an irrational number between the two, and we can use that rational number to create a disconnection. Therefore the only connected subsets of $\mathbb{Q}$ are the singleton sets $\{q\}$, where $q \in \mathbb{Q}$, which are closed.       We will show that $f(\mathbb{R})$ must be connected. Assume not, then    $$ f(\mathbb{R})= U\cup V, $$   where $U$ and $V$ are nonempty open subsets of $\mathbb{Q}$ such that    $$ U \cap V= \emptyset .$$    This implies   $$ \mathbb{R} = f^{-1} \big( U \cup V \big) = f^{-1}(U)\cup f^{-1}(V),  $$   where $f^{-1}(U),f^{-1}(V)$ are nonempty, nonintersecting subsets of $\mathbb{R}$. This, however, is a contradiction, as $\mathbb{R}$ is connected. Therefore no such set $U$ and $V$ can exist.       As we have already shown, the only connected subsets of $\mathbb{Q}$ are the singleton sets $\{q\}$, so $f(\mathbb{R})$ must be such a set.   Something about this does not seem quite satisfactory, as if I am missing something. Could anyone tell me a flaw in my logic? Also, is there a more satisfactory way to prove this theorem?
31488
In :  $$ \begin{align} t(x) = \begin{cases} 0 &amp; \text{if $x$ is irrational}\\ \frac{1}{n} &amp; \text{if $x = \frac{m}{n}$ where $\gcd(m,n) = 1$} \end{cases} \end{align} $$  I can prove the discontinuity at rational $b$ by taking a sequence of irrationals $x_n$ which converge to $b$.   But while going through an argument for continuity at irrationals.  I found this in a book.      On the other hand if $b$ is an irrational number and $\epsilon &gt; 0$   then there is a     natural number $n_0$ such that $1/n_0 &lt; \epsilon$.   There are only finite number of rationals with denominator less than   $n_0$ in the interval $(b-1,b+1)$. Hence we can find a $\delta &gt; 0$   such that $\delta$ neighbourhood of $b$ contains no rational with   denominator less than $n_0$.   I understand the rest of the proof. But I am unable to prove the emphasized text. Although I find it intuitive.
85465
Possible Duplicate:          I think I saw somewhere(but I'm not sure) that there is no function $g$ on $[0,1]$ that is continuous at all rational points and discontinuous at all irrational points. Please, is this true? If yes how can I show it?  Thanks.
21944
Pretty much the problem asks if a function is continuous at the point $c$ and $f(c) &gt; 0$ then there exists a $d &gt; 0$ such that $\forall x$, $f(x) &gt; 0$ with $|x-c| &lt; d$.   I can understand what the problem means. That if a function is positive at a point then there is another point that's really close that'll also be positive. I can not prove this in a formal way though. I've tried using the intermediate value theorem and I do not know hot to implement it. Any ideas?
8495
I have a question, to aid my understanding, about the proof that differentiability implies continutity.$\mathstrut$    Differentiability Definition  When we say a function is differentiable at $x_0$, we mean that the limit:  $$‎f^{\prime} ‎(x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists.  Continuity Definition  When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$  Theorem: Differentiability implies Continuity: If $f$ is a differentiable function at $x_0$, then it is continuous at $x_0$.  Proof:  Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} =  ‎f^{\prime} ‎(x) $$  and hence  $$ \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0$$  We have therefore shown that, using the definition of continuous, if the function is differentiable at $x_0$, it must also be continuous.  My Question  The proof seems to execute the following steps:   Assume the function is continuous at $x_0$ Show that, with little algebra, we can change this into an equivalent question about differentiability at $x_0$. With this little bit of algebra, we can show that if a function is differentiable at $x_0$ it is also continuous.   What I am slightly unsure about is the apparent circularity. In my mind it seems to say, if a function is continuous, we can show that if it is also differentiable, then it is continuous. Rather than what I was expecting, namely, if a function is differentiable, we can show it must be continuous.  Hopefully my confusion is clear. Any help will be greatly appreciated.
2015
Suppose that $f$ is continuous at $x_0$ and $f$ satisfies $f(x)+f(y)=f(x+y)$. Then how can we prove that $f$ is continuous at $x$ for all $x$? I seems to have problem doing anything with it. Thanks in advance.
9608
Let $f\colon[0,1]\to\mathbb{R}$ be continuous such that $f(x)\in\mathbb{Q}$ for any $x\in[0,1]$. Intuitively I feel that $f$ is constant, since $\mathbb{Q}$ is dense in $\mathbb{R}$.  How can I formally write this down?
39807
We know that the points of discontinuity of a monotone function on an interval $[a,b]$ are countable. Using this can we prov that:    Any bijection $f: \mathbb{R} \to [0,\infty)$ has infinitely many points of discontinuity.   If yes, how or otherwise how to prove the above result?
102312
This is question B4 in 1999 Putnam Exam. I would appreciate if somebody could help me with this.      Let $f(x),f'(x),f''(x),f'''(x)&gt;0$ , $f'''(x)$ is a continuous function and $f'''(x)&lt;f(x)$ on $\mathbb{R}$. Then show that   $$f'(x)&lt;2f(x),~ \forall x\in \mathbb{R}.$$
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Akhil showed that the  is the same as the continuum, using as a step the observation that continuous functions that agree at rational points must agree everywhere, since the rationals are dense in the reals.  This isn't an obvious step, so why is it true?
95337
Possible Duplicate:          Let $f, g:\Bbb{R}\to\Bbb{R}$ to be continuous functions such that $f(x)=g(x)\text{ for all rational numbers}\,x\in\Bbb{Q}$. Does it follow that $f(x)=g(x)$ for all real numbers $x$?  Here is what I think:  f continuous when $\lim\limits_{x\to x_0}f(x)=f(x_0)$ and $\lim\limits_{x\to x_0}g(x)=g(x_0)$  So it does not neccesarily mean that $f(x)=g(x)$ when x is irrational. So I can pick a function f so that   $f(x) = \begin{cases} g(x)  &amp; \text{if $x\in\Bbb{Q}$} \\ x &amp; \text{if $x\in\Bbb{R}\setminus \Bbb{Q}$}  \\ \end{cases}   $
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I know there are many answers for this question. I had read almost every answer And try get arguments But something that goes missing in each time. I had been stuck on this problem since half day but not understating. Please help me in this regard. Let $f$ be defined on $[0,1]$;  $$ \begin{align} f(x) = \begin{cases} 0 &amp; \text{if $x$ is irrational}\\ \frac{1}{q} &amp; \text{if $x = \frac{p}{q}$ where $(p,q) = 1$ and $q$ &gt; 0}. \end{cases} \end{align} $$ For the rational numbers, I consider a sequence of irrational points $x_n$ converging to some rational number say $m/n$.  If this function is continuous for any rational say $m/n$ then this must hold: $\forall \epsilon &gt;0$ ,$\exists \delta&gt;0$ such that $|1/n|&lt;\epsilon $ whenever $|x_n-m/n|&lt;\delta$.But for any particular $m/n,1/n$ need not be small. Therefore function is discontinuous at the rational points.  For the irrational points, consider rational sequence $y_n$ converging to $a$, irrational number. Consider $y_n \to a$ therefore $\forall \epsilon &gt;0 $, $\exists N$ such that $|y_n-a|&lt;\epsilon $ $\forall n&gt;N$ In case the function is continuous at $a$ then it must satisfy following $\forall \epsilon &gt;0 $ ,$\exists \delta&gt;0$ such that|$f(y_n)$|&lt;$\epsilon $ whenever $|y_n-a|&lt;\delta$. I wanted to make that $f(y_n)$ small for that $y_n$ in neighborhood  of $a$.From this I am not able convince myself from last answers.   This is not a duplicate, since I specified my particular problem with this proof. Please help me.
64935
How would you prove or disprove that the function given by $$f(x,y) = \begin{cases} \dfrac{x^2y}{x^2 + y^2} &amp; (x,y) \neq (0,0) \\ 0 &amp; (x,y) = (0,0) \end{cases}$$ is continuous at $(0,0$)?
77224
Let $f:\mathbb{R}\to\mathbb{R}$ , there is $k \in \mathbb{R}$, $|k|\ne 1$, so that for every $x\in \mathbb{R} , f(kx)=f(x).$ Show that if $f$ continuous at $0$ then $f$ is a constant function.   I tried to find something to show it from the definition of continuous function but I didn't know how to continue. We have that $f$ continuous at $0$ if for every $\epsilon&gt;0$ there is $\delta&gt;0$ such that $|x-0|&lt;\delta \implies |f(x)-f(0)|&lt;\epsilon$ And if $f$ not continuous on $0$ is $f$ still a constant function?
77549
Let $f:[0,1]\to\mathbb R$ is a continuous function, and for all $x\in [0,1]$, $f(x)$ is either a local maximum or a local minimum. Then prove that $f$ is a constant.  Here is what I have tried, I don't know whether it is correct:  Assume $f$ is not a constant, by continuity, there exist $x_1$, $x_2$ such that $f(x_1)=m$ is a global minimum and $f(x_2)=M$ is a global maximum. Now if $m\ne M$, choose any $c\in(m,M)$ and define  $$x_0=\sup (x\in[x_1,x_2]:f(x)&lt;c)$$  Then we have $f(x_0)$ NOT a local minimum nor a local maximum (since according to the definition of $x_0$, there are always some points $&lt;c$ in the left neighborhood of $x_0$ and the points in right neighborhood are always $&gt;c$). Contradiction arises and so $f$ must be a constant.