Let $\langle r_n\rangle$ be an enumeration of the set $\mathbb Q$ of rational numbers such that $r_n \neq r_m\,$ if $\,n\neq m.$ $$\text{Define}\; f: \mathbb R \to \mathbb R\;\text{by}\;\displaystyle f(x) = \sum_{r_n \leq x} 1/n^{2},\;x\in \mathbb R.$$   Prove that $f$ is continuous at each point of $\mathbb Q^c$ and discontinuous at each point of $\mathbb Q$.   I find it difficult to prove especially the continuity on irrationals, I proved the discontinuity on a rational number in the following way is it correct?  Let $ c \in \Bbb Q $ then $ c=r_n $ for some $n \in \Bbb N  $ and  Let $ \epsilon_0 = 1/n^{2} $  Let $\delta &gt; 0 $ be arbitrary and let $ x \in (c-\delta. c+\delta)$ such that $x&lt;c $   Then $|f(x)-f(c)|&gt;1/n^{2}=\epsilon_0 $  How do prove that it is continuous on irrationals?
123744
Sorry, I was having trouble giving the question because I can't figure out how to type in mathematical symbols. Let me try again:  $\{q\}$ is an enumeration of the rational numbers, and   $$f(x)=\sum_{q_n&lt; x}\frac1{n^2}$$  for $x\in\Bbb R$.  Or $f(x)=\sum(1/n^2)$, the index of summation $q_n&lt; x$, and $q$ is an enumeration of the rational numbers.  The goal is to prove that $f$ is continuous at each irrational and discontinuous at each irrational.  I'm having trouble visualizing this series so I'm not sure why it is supposed to be different at irrational as opposed to a rational $x$ ...  I think I need to use the epislon-delta definition of continuity but I'm not sure what epsilon to use.
22355
From Pugh's analysis book, prelim problem 57 from Chapter 4:     Let $f$ and $f_n$ be functions from $\Bbb R$ to $\Bbb R$. Assume that $f_n(x_n)\to f(x)$ as $n\to\infty$ whenever $x_n\to x$. Prove that $f$ is continuous. (Note: the functions $f_n$ are not assumed to be continuous.)   here's my attempt: assume $x_n \to x$. we want to show that $f(x_n) \to f(x)$. so $|f(x_n) - f(x)| \leq |f(x_n)-f_n(x_n)| + |f_n(x_n)-f(x)|$. The second term can be made to be less than any $\varepsilon &gt; 0$ for $n$ sufficiently large. i'm having trouble with the first term. can anyone help? thank you!
102678
Prove that any continuous map on $\mathbb{R}$ which is the identity on $\mathbb{Q}$ is the identity map, hence Aut$(\mathbb{R}/\mathbb{Q})= 1.$  proof: Suppose $\sigma$ is continuous on $\mathbb{R}$ which is an identity on $\mathbb{Q}$. let $a \in R $. Then since $\sigma$ is continuous, then by definition there is $\delta_1 &gt; 0$ such that $|a - b| &lt; \delta_1$.  And assume $|\sigma(a) -\sigma(b)| &lt; \epsilon/2$ .  Let $\delta $= min{$\epsilon/2, \delta_1$}.  And now let $b \in \mathbb{Q}$ such that $|a - b| &lt; \epsilon$. Then  $|\sigma(a) - a| = |\sigma(a)  - b + b - a| \leq |\sigma(a) - \sigma(b) | + |b - a| &lt; \epsilon/2 + |b-a|$.  Can someone please help me? I am not sure . Can I say $|b - a| &lt; \epsilon/2 $?
18196
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!
53477
Suppose $\epsilon &gt; 0$ and that $c \in \mathbb{R}$. Since $f^3$ is continuous, we can choose $\delta &gt; 0$ such that whenever $\lvert x - c \rvert &lt; \delta$ we have that $\rvert f^3(x) - f^3(c) \lvert &lt; \epsilon$.   To show that $f$ is continuous, I know I need some number $\delta_1 &gt; 0$ such that $\lvert x - c \rvert &lt; \delta_1$ implies that $\lvert f(x) - f(c) \rvert &lt; \epsilon$.   I've tried manipulating the expression $\lvert f(x) - f(c) \rvert &lt; \epsilon$ by using the fact that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. I get  $$   \lvert f(x) - f(c) \rvert &lt; \frac{\lvert f^3(x) - f^3(c)\rvert}{\lvert f^2(x) + f(x)f(c) + f^2(c) \rvert} &lt; \frac{\lvert f^3(x) - f^3(c)\rvert}{\lvert f^2(x) + f(x)f(c) \rvert}. $$ However, I am not sure how to continue and I am also not confident that this is a good way to approach the exercise.
11185
Suppose that $f : [a, b] \times [c, d] \to\mathbb R$ is a continuous function. Let $$G(y)= \int_a^b f(x, y) \, dx$$ $$H(x)= \int_a^b f(x, y) \, dy$$  Prove that $G$  is continuous on $[c, d]$ and $H$ is continuous on $[a, b]$.  So I know that the function will be uniformly continuous on the interval but I don't know where to go from there
1866
Let f,g be continuous maps of S in $\mathbb{R}^n$, to $\mathbb{R}^m$. Show the inner product h(x) = $\langle f(x), g(x)\rangle$ is continuous.  My attempt: Since f,g are continuous, their product is a continuous mapping, so for $\epsilon \gt 0\: t.e\:  \delta \gt 0$ such that $\parallel x - a\parallel \lt \delta $ implies $\parallel \langle f(x),g(x)\rangle - \langle f(a)g(a)\rangle \parallel \le\parallel f(x)g(x) - f(a)g(a)\parallel \lt \epsilon$ by schwartz inequality.
54565
If $f$ is an increasing function defined on $[a,b]$ . Then show that the set of discontinuities of $f$ is countable and that $f$ has points of continuity in every open subinterval of $[a,b]$  Attempt: I have trouble understanding why the number of discontinuities of $f$ should be countable. For example, if we define a function which is increasing but different at each point in the interval $[0,1]$, then, there can be uncountable number of discontinuities.  I think if we prove that the number of discontinuities is countable, then we can prove that the function has points of continuity in every open interval of $[a,b]$.  Please help me understand intuitively why $f$ must necessarily have countable number of discontinuities. Suppose, $f$ has an uncountable number of discontinuities, then what can happen?  Thank you for your help.
77194
$\newcommand{\R}{\mathbb R}$  Let $f: \R \to \R$ be an increasing function. $(a \leq b \Rightarrow f(a) \leq f(b))$. I want to prove that $f \in \mathcal B$, the Baire one functions.  I first tried to prove that $g(x) := \lfloor f(x)+x \rfloor$ is in $\mathcal B$. The points of discontinuity of $g$ are countable since $g$ is increasing. Now, let $\mathcal D$ be the points in $\R$ where $g$ is not continuous. If $z \in \mathcal D$ then I want to prove that $\exists \delta &gt; 0: (z-\delta,z+\delta) \cap \mathcal D = \{z\}$, i.e. $z$ is the only discontinuity in a certain neighborhood. Then it is not hard to prove that $g \in \mathcal B$ and that $f$ is the uniform limit of $\mathcal B$ functions and thus is also $\mathcal B$.  I already tried to prove this (existence of $\delta &gt; 0$) by contradiction. However, I could not see a nice way to get here a contradiction.
106948
Prove that if $f, g$ : $X$ → $\mathbb R$ are continuous at $a$ ∈ $\mathbb R$, then $f · g$ is continuous in $a$.  If $f$ is continuous at $a$, then $∀ε_f &gt; 0, ∃δ_f &gt; 0$ such that  $|x-a| &lt; δ_f$ iff $|f(x) - f(a)| &lt; ε_f$  and if $g$ is continuous at $a$, then $∀ε_g &gt; 0, ∃δ_g &gt; 0$ such that  $|x-a| &lt; δ_g$ iff $|g(x) - g(a)| &lt; ε_g$.  Now make $|f(x)g(x) - f(a)g(a)|$ =  $|f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a)|$ $≤$  $|f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)|$ &lt; $ε_f|g(a)| + ε_g|f(x)|$. (1)  Notice that if $|f(x) - f(a)| &lt; ε_f$ then $|f(x)| - |f(a)| &lt; ε_f$, so $|f(x)| &lt; ε_f + |f(a)|$.  That implies, from (1): $|f(x)g(x) - f(a)g(a)| &lt; ε_f|g(a)| + ε_g(ε_f + |f(a)|) = ε_f(|g(a)| + ε_g|f(a)|)$.  Now I'm a little lost. From the proofs I've seen, it seems to me that I could simply take $δ = min(δ_g, δ_f)$. Also, since $ε_f$ and $ε_g$ can be made as small as one wants, there will be some δ that satisfies the conditions for continuity. But from what I read, I should explicitly show a δ in terms of ε. Anyway, I reasoned that since $f$ and $g$ are continuous, I can make bounds on $x$ symmetrically* as done to $f(x)g(x)$ and got the following:  δ = $δ_f(|a| + δ_g|a|)$ = $δ_f|a|(1 + δ_g)$  so:  $|x-a| &lt; δ_f|a|(1 + δ_g)$ iff $|f(x)g(x) - f(a)g(a)| &lt;  ε_f(|g(a)| + ε_g|f(a)|)$   I need to clarify/better define this, but it is basically the notion that making the same operations I made on the bounds of $f(x)g(x)$ in relation to the bounds of $f(x)$ and $g(x)$ individually over the bounds of $x$ for each function, should leave me with the adequate bounds for $x$ on the composite function.
121446
Related;    This is the theorem i need to prove;  "Let $E(\subset \mathbb{R})$ be closed subset and $f:E\rightarrow \mathbb{R}$ be a contiuous function. Then there exists a continuous function $g:\mathbb{R} \rightarrow \mathbb{R}$ such that $g(x)=f(x), \forall x\in E$."  I have tried hours to prove this, but couldn't. I found some solutions, but ridiculously all are wrong. Every solution states that "If $x\in E$ and $x$ is not an interior point of $E$, then $x$ is an endpoint of a segment of at most countable collection of disjoint segments.". However, this is indeed false! (Check Arthur's argument in the link above)  Wrong solution Q4.5;    Just like the argument in this solution, i can see that $g$ is continuous on $E^c$ and $Int(E)$. But how do i show that $g$ is continuous on $E$?
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 ?
40221
Let $h:[0,1]\to\mathbb R$  $h(x)=\begin{cases}0&amp;\text{if }x=1\\\frac{1}{n}&amp; \text{otherwise if }x\in\mathbb Q,x=\frac{m}{n},\;m,n\in\mathbb N,\gcd(m,n)=1\\0&amp;\text{otherwise if }x\in\mathbb R\setminus\mathbb Q\end{cases}$  How do you prove that $h$ is continuous on all irrational points within $[0,1]$?
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?
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?
61445
I was trying to prove an important theorem concerning continuous functions, namely that the product of two continuous functions is continuous. I am nearly at the end of the proof but I do not understand the last step my teacher made... I'm sure some of you is able to help me out :) Thanks in advance!  Theorem  Let $f, g\colon \Bbb R \to\Bbb R $ be continuous. Then $F\colon \Bbb R\to\Bbb R$ defined by $F(x) = f(x)g(x)$ is continuous.  Proof  Let $f, g\colon \Bbb R \to\Bbb R $ be given such that $f$ and $g$ are continuous.  Let $F\colon \Bbb R\to\Bbb R$ be defined by $F(x) = f(x)g(x)$. We want to show: $F$ is continuous, that is, for all $a \in\Bbb R$, for every $\epsilon &gt; 0$, there exists some $\delta &gt; 0$ such that for all $x \in \Bbb R$ with $|x - a| &lt; \delta$, $|F(x)-F(a)| &lt; \epsilon$.  Let $a \in\Bbb R$ be given. Let $\epsilon &gt; 0$ be given.   Let $\delta_f &gt; 0$ be such that for all $x \in\Bbb R$ with $|x-a| &lt; \delta_f$, $|f(x)-f(a)| &lt; \frac\epsilon{2(|g(a)| + \epsilon)}$. Such a $\delta_f$ exists since $f$ is continuous.  Let $\delta_g &gt; 0$ be such that for all $x \in\Bbb R$ with $|x-a| &lt; \delta_g$, $|g(x)-g(a)| &lt; \frac\epsilon{ 2|f(a)| + 1}$. Such a $\delta_g$ exists since $g$ is continuous.  Let $\delta_3 &gt; 0$ be such that for all $x\in\Bbb R$ with $|x-a| &lt; \delta_3$, $|g(x)-g(a)| &lt; \epsilon$, which implies $|g(x)| &lt; |g(a)| + \epsilon$. Such a $\delta_3$ exists since $g$ is continuous.  Choose $\delta = \min\{\delta_f, \delta_g, \delta_3\}$.  Then for all $x \in\Bbb R$ with $|x-a| &lt; \delta$:  $$\begin{align}|F(x) - F(a)| &amp;= |f(x)g(x) - f(a)g(a)| \\ &amp;= |f(x)g(x) - f(a)g(x) + f(a)g(x) - f(a)g(a)| \\ &amp;\le |f(x)g(x) - f(a)g(x)| + |f(a)g(x) - f(a)g(a)|\\ &amp;= |(g(x)[f(x) - f(a)]| + |f(a)[g(x) - g(a)]|\\&amp;= |g(x)| * |f(x)-f(a)| + |f(a)| * |g(x) - g(a)| \\ &amp;&lt; [|g(x)| + \epsilon] \cdot \frac \epsilon {2(|g(a)| + \epsilon)} + |f(a)| \cdot |g(x) - g(a)| \\ &amp;= \frac\epsilon{2 + |f(a)| \cdot |g(x) - g(a)|}\end{align}$$  Problem  Now we need $|f(a)| \cdot |g(x) - g(a)|$ to be equal to $\frac \epsilon 2$ so that we have $\frac\epsilon2+\frac\epsilon 2= \epsilon$ (which completes the proof). The problem is I do not see why/how $|f(a)| \cdot |g(x) - g(a)|$ is equal to $\frac\epsilon2$. Or is there maybe an error somewhere?
31866
How can I resolve the task below:  Given $f(x)= \begin{cases}   x, &amp;x\in \mathbb{Q}\text{  }\\   1-x, &amp;x\notin \mathbb{Q}\text{ (irrational)}  \end{cases}$, $0 \leq x \leq 1$.  Show $f(x)$ is continuous only in $x=\dfrac{1}{2}$.