I am trying to read Mathematical Analysis I by Zorich on my own. Here is an exercise that I could not solve:      For all $l \in \mathbb{R}$ that is not of the form $\frac{1}{n}$ for some $n \in \mathbb{N}$, there exists a continuous function $f: [0, 1] \Rightarrow \mathbb{R}$ such that $f(0) = f(1)$ and the graph of $f$ contains no horizontal chord with length $l$.    I was wondering how to prove this? Thanks in advance.
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I've been reading through Zorich's "Analysis I" book recently, and I came across this nice little exercise.  Let $f: [0,1]\to \mathbb R$ be a continuous function such that $f(0)=f(1)$. Show that    for any $n\in \mathbb N$ there exists a horizontal closed interval of length $\frac 1n$ with endpoints on the graph of this function; if the number $\ell$ is not of the form $\frac 1n$ there exists a function of this form on whose graph one cannot inscribe a horizontal chord of length $\ell$.   The first part can be proven like this: Consider $g: [0,(n-1)/n] \to \mathbb R$ given by $g(x) = f(x) - f(x+1/n)$. Then   $$\sum_{k=0}^{n-1} g(k/n) = 0$$  and therefore either all of these points are zero or there exists both a point where $g$ is positive and a point where $g$ is negative. By continuity, there must then also be a point where $g = 0$. So we are done.  Now, the second statement seemed rather counterintuitive, and I have given it some time now, but don't see a counterexample for $\ell &lt; 1/2$.  (For $\ell &gt; 1/2$ the function $f(x) = \sin(2\pi x)$ will do.)  Can anyone help me out?  Cheers,
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Given a continuous function $f:\mathbb{Q}\to\mathbb{Q}$ ,does there exist a continuous function  $g:\mathbb{R}\to\mathbb{R}$ such that   $g|_{\Bbb Q} = f$?  What I have no Idea about how to attempt this Question! Any suggestion will be very helpful.
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Another question I am struggling with:  Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $\lim_{x \rightarrow a}\left(f(x)+\dfrac{1}{\left|f(x)\right|}\right)=0$. Find $\lim_{x \rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.    Thank you all for your answers!!  But.. How can I prove by definition that this is indeed the limit? take $\epsilon&gt;0$, I have to show that there exists a $\delta&gt;0$ such that, $|x-a| \leq \delta$ implies $|f(x)+1| \leq \epsilon$. now, I do know that there exists an $\delta_1&gt;0$, such that $|f(x)+\frac{1}{|f(x)|}|&lt; \epsilon$.Or, $\frac{\epsilon}{k}$ for some constant $k$ But, how can I continue from there?
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I need to show that if $f$ is continuous function ($f:\mathbb{R}\rightarrow \mathbb{R}$), then there exists a sequence of polynomials which converges to $f$ on any compact subset of $\mathbb{R}$.   I would appreciate any help/direction ...
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Which of the following is a continuous image of $GL_n(\mathbb{R})$?  The real line $\mathbb{R}$  The subspace $\{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$ of $\mathbb{R}^2$  $\mathbb{R}^2 \setminus \{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$  $\{(x_1,x_2,...,x_n):x_1^2+x_2^2+...+x_n^2 \le 1\}$ of $\mathbb{R}^n$   How do I proceed? $GL_n(\mathbb{R})$ is not compact, not connected, so I cannot rule out some options using topological properties.
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This is the starred example 5.1.8 in Krantz' Real Analysis and Foundations.     Give an example of a function $f: \mathbb{R} \rightarrow \mathbb{R}$ so that $\lim_{x\rightarrow c} f(x)$ exists when $c$ is irrational but does not exist when $c$ is rational.   I'm aware of the Dirichlet and Thomae functions but this seems slightly different. I'm not even sure a solution exists. Thanks for any help!
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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?
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Let $f$ be a function defined on $[0,1]$ by   $$f(x) = { 0, \text{ if } x = 0} $$ $$f(x) = { x \sin \frac 1 x , \text{ if } 0 &lt; x \leq 1} $$  Prove that the curve $\{(x, f(x)) : x \in [0,1]\}$ is not rectifiable.   I'm not sure how to approach this. The general idea seems logical, we're proving that the length of the curve is infinite, but the method seems difficult to find.
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I'm wondering if it's possible to have a continuous sequence $f: \mathbb{N} \to \mathbb{R}$? My intuition is telling me no because logically it would be impossible to map the natural numbers onto the real numbers but I'm not sure how to formalize it. What do you think? How would you prove that no function over the natural numbers (no sequence) is continuous?  For example, how would you prove that $f: \mathbb{N} \to \mathbb{R}$ such that $f(n) = n^{2}$ is not continuous over the natural numbers?
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If $f:[0,1] \rightarrow \mathbb{R}$ is a continuous function such that $f(0)=f(1)$, then there exists $ x\in [0,1]$ such that  $f(x) = f(x+\frac{1}{n})$, where $n$ is any natural number.  Let $g(x)=f(x+\frac{1}{n})-f(x)$ on $[0,1-\frac{1}{n}]$. Now, $f$ is continuous function on closed interval.So, it must attain its bounds. Let $m$ be minima at $c_1 \in [0,1]$ and $M$ be maxima at $c_2 \in [0,1]$.  Now if $c_1,c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $g(x)$ on $[c_1,c_2]$.  If $c_1$ or $c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $[c_1,c_1-\frac{1}{n}]$ or  $[c_2,c_2-\frac{1}{n}]$ respectively. If  $c_1,c_2$ both do not belong to $[0,1-\frac{1}{n}]$, then apply IVT to $g$ on $[c_1-\frac{1}{n},c_2-\frac{1}{n}]$.  Is this proof correct? Thanks.
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Is it possible to find a continuous function $f$: $[0,1]$ $\rightarrow$ {$0,1$} that is onto?  My thoughts: I know if $A$ &amp; $B$ are two subsets of a top. space $X$ and if there exists a continuous function $f$: $X$ $\rightarrow [0,1]$ s.t $f(A)=${0} &amp; $f(B)=${1}, we say $A$ &amp; $B$ can be a separated by a continuous function $f$.  But the problem is $[0,1]$ is connected and hence can't find a separation of it.Is it possible to find such a function?.   Alternatively, I know Urysohn lemma and the definition of a completely regular space(which are somewhat related to this but not quite). Any thoughts?
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I'm au courant with  and . They're both Abbott P111 exm 4.3.6 which proves "a continuous function is sometimes described, intuitively, as one whose graph can be drawn $\color{seagreen}{by \; dint \; of}$  without lifting your pencil from the paper. I countenance this.  But I don't think the underneath is a counterexample? I see $f(n) = n$ for all natural numbers $n$ is discontinuous over $\mathbb{R} \times \mathbb{R}$. But what if I visualize over $\mathbb{N} \times \mathbb{N}$? Then $f(n) = n$ can be drawn without lifting your pencil ?    (p 1 sur 14 )
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Let $f:R^n \rightarrow R^m$ be any function. Will the graph of $f$ always have Lebesgue measure zero?  $(1)$ I could prove that this is true if $f$ is continuous.  $(2)$ I suspect it is true if $f$ is measurable, but I'm not sure. (My idea was to use Fubini's Theorem to integrate the indicator function of the graph, but I don't know if I'm using the Theorem properly).  If $(2)$ is incorrect, what would be a counterexample where the graph of $f$ has positive measure?  If $(2)$ is correct, can we prove the existence of a non-measurable function whose graph has positive outer measure?
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How many continuous function $f(x)$ exist such that $$\int_{0}^{1}f(x)\big(1-f(x)\big)\mathrm dx = \frac{1}{4}\text?$$ I want to find the number of such functions but I don't know how to proceed. It would be of great help if someone can tell me which theorem to apply or provide with any other hint. Any help is highly appreciated.
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Let $f$ and $g$ be continuous functions and map from $[0,1]$ to $[0,1]$. Also let $f(g(x)) = g(f(x))$ . Prove that there exists $c$ from $[0,1]$ such that $f(c)=g(c)$.   I will try with contradiction. Let $h(x) = f(x) - g(x) &gt; 0$ for all $x$ from $[0,1]$. Since $f(x)$ maps from $[0,1]$ to $[0,1]$ and is greater then $g$ for all $x$ from $[0,1]$ then this implies that $g(x)$ is not an element of $[0,1]$ for all $x$ from $[0,1]$. This is a contradiction, so there exists $c$ from $[0,1]$ such that $f(c)=g(c)$. Problem here is that I never used fact that  $f(g(x)) = g(f(x))$ which bothers me. Is proof correct or there is hole somewhere?
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So the question asks to find a topology on $\mathbb{R}$ so that all the polynomials are continuous functions from $\mathbb{R}$ to $\mathbb{R}$ but $\cos(x)$ is not continuous as a function from $\mathbb{R}$ to $\mathbb{R}$.   I looked at another similar question in stack exchange:  I think this is a special case of my question since it requires all other functions to not be continuous.   Other than that I thought maybe the solution could be based on how cosine is a periodic function unlike the polynomials, so maybe I should take out some infinite sets from the topology but that still wouldn't be enough since the topology would be closed under unions. So I am not even sure if there exists such topology.
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Determine all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x, y \in \mathbb{R}$, $$f((x + 1) f(y)) = y (f(x) + 1).$$    Source: Vuong Lam Huy, on a Facebook group.