Let $f:\mathbb{R}\to \mathbb{R}$ and denote $A_f=\{(x,f(x)),x\in\mathbb{R}\}$.  I've proved that if $f$ is continuous, then $A_f$ is closed. But I have no clue for the other direction. Does $A_f$ closed also imply $f$ continuous?  Thanks a lot~
79821
I tried looking up a question regarding graphs of continuous functions on this site, but all the ones I found consider functions from $\mathbb{R}$ into $\mathbb{R}$. I have been pondering the following question: given a general topological spaces $X, Y$, and a function $f: X\to Y$, when does $Graph(f)$ closed in $X\times Y$ imply that $f$ is continuous. By the closed graph theorem, this is true whenever $X$ and $Y$ are both Banach spaces.   Also, it is fairly easy to prove that whenever $Y$ is a Hausdorff space and $f$ is continuous, then $Graph(f)$ is closed, but I do not think that the converse is true, so I am trying to find an example where $X$ is some topological space, $Y$ a Hausdorff space, $f: X\to Y$ a function with a closed graph in $X\times Y$, but who fails to be continuous. As of yet I have not been able to find such a counterexample, partially because I have no clue where to look for such a counterexample. I would really appreciate getting some directions to go in.
79893
I wanna prove that   "if $f: \mathbb{R}^n \to \mathbb{R}$ is continuous and satisfies $f=0$ almost everywhere (in the sense of Lebesgue measure), then, $f=0$ everywhere."  I am confident that the statement is true, but stuck with the proof. Also, is the statement true if the domain $\mathbb{R}^n$ is restricted to $\Omega \subseteq \mathbb{R}^n$ that contains a neighborhood of the origin "$0$"?
49710
Suppose $Y$ be an ordered set in the order topology. Let $f, g: X \to Y$ be continuous. How to show that the set $\{x| f(x) \leq g(x)\}$ is closed?  This is a excercise from munkres. Maybe trying to show that the set $f(\{x| f(x) &gt; g(x)\})$ is open is suffice. But I could not figure out the connection between this set and the continuity of the two functions.Could you give a hint?  Thanks.
44936
Suppose $f:[0,1)\to (0,1)$ is bijective. Prove that $f$ is not continuous.     I know that $(0,1)$ and $[0,1)$ are not homeomorphic spaces because of the connected property. Can we then conclude that $f$ is not continuous because of that?
121926
A function $f:\Bbb R \to \Bbb R$ is additive and Lebesgue measurable. Prove that $f$ is continuous.  I know that on $\Bbb Q$, $f$ comes out to be linear. So, if $f$ is to be continuous then $f$ must be linear in $\Bbb R$. But, I'm stuck here.  If anyone can please help. Thanks in advance.
95875
I got started recently on proofs about continuity and so on. So to start working with this on $n$-spaces I've selected to prove that every linear function $f: \mathbb{R}^n \to \mathbb{R}^m$ is continuous at every $a \in \mathbb{R}^n$. Since I'm just getting started with this kind of proof I just want to know if my proof is okay or if there's any inconsistency. My proof is as follows:  Since $f$ is linear, we know that there's some $k\in \mathbb{R}$ such that $|f(x)|\leq k|x|$ for every $x\in \mathbb{R}^n$, in that case let $a\in \mathbb{R}^n$ and let $\varepsilon &gt;0$. Consider $\delta = \varepsilon /k$ and suppose $|x-a|&lt;\delta$, in that case we have:  $$|f(x)-f(a)|=|f(x-a)|\leq k |x-a|&lt;k \frac{\varepsilon}{k}=\varepsilon$$  And since $|x-a|&lt;\delta$ implies $|f(x)-f(a)|&lt;\varepsilon$ we have that $f$ is continuous at $a \in \mathbb{R}^n$. Since $a$ was arbitrary, $f$ is continous in $\mathbb{R}^n$. Is this proof fine? Or there was something I've missed on the way?
19321
I am thinking about this since couple hour. Is a continuous and bounded function $f:\mathbb R\to\mathbb R$ uniform continuous too? I didn't found a counter example and thus I tried to prove this like this: let $\epsilon&gt;0$ and $a\in\mathbb R$. By continuity, there is a $\delta_a&gt;0$ such that $$d(f(x),f(y))&lt;d(f(x),f(a))+d(f(y),f(a))&lt;\epsilon$$ if $d(x,y)\leq d(x,a)+d(y,a)&lt;\delta_a.$ Then, if $\inf_a\delta_a&gt;0$ I set $\delta=\inf_a\delta_a$ and we are done since $d(f(x),f(y))&lt;\epsilon$ if $d(x,y)&lt;\delta,$ but I didn't success to prove that $\delta\neq 0$.
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 ?
119882
I came across a problem that says:  Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function. If $|f'|$ is bounded, then which of the following option(s) is/are true?     (a) The function $f$ is bounded.   (b) The limit $\lim_{x\to\infty}f(x)$ exists.   (c) The function $f$ is uniformly continuous.   (d) The set $\{x \mid f(x)=0\}$ is compact.   I am stuck on this problem. Please help. Thanks in advance for your time.
23925
If $ f: \mathbb{R} \to \mathbb{R} $ is continuous, and $\forall x \in \mathbb{R} :\;(f \circ f \circ f)(x) = x  $, show that $ f(x) = x $.  The condition that $f$ is continuous on $\mathbb{R}$ is crucial for the proof. I can find functions such as $\displaystyle\frac{x-3}{x+1}$ that satisfies  $ (f \circ f \circ f)(x) = x $. I have tried to negate the conclusion to see if there's a contradiction, but got stuck.
40237
Suppose $f(x)$ is continuous on the closed interval $[a,b]$. Define $m(x)=\max_{a\leq s\leq x}\, f(s)$, $a\leq x\leq b$. Is $m(x)$ continuous necessarily? Thank you.
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!
27045
I was just trying to solve an exercise in Munkres' Analysis on Manifolds to get better at continuous functions and topology in $\mathbb{R}^{n}$, and I got stuck. Here it is:      Let $X = A \cup B$, $A$ and $B$ are subspaces of $X$. Let $f:X \rightarrow Y.$ Suppose that the restricted functions $f_{|A}:A \rightarrow Y$ and $f_{|B}:B \rightarrow Y$ are continuous. Show that if both $A$ and $B$ are closed in $X$, then $f$ is continuous.    Our definition of continuity is:  $f$ is continuous at $x \in X$ if for each open set $V$ of $Y$ containing $f(x)$, there is an open set $U$ of $X$ containing $x$ such that $f(U) \subset V$. $f$ is continuous if it is continuous at each point of its domain.  What I've tried so far:  Suppose $A$ and $B$ are closed in $X$. Suppose $A \cap B = \emptyset$. Then $X -A$ is open in $X$. But $X - A = B$ which is closed in $X$, which is a contradiction. Hence, $A \cap B \ne  \emptyset$. I'm not sure if this will be useful in the problem or not. If $x \in X$, then it's either in $A$ or in $B$. Suppose without loss of generality that it's in $A$. Then for all open sets $V$ of $Y$ s.t. $f(x) \in V$, $\exists$ an open set $U_{1}$ of $A$ s.t. $U_{1} \subset A, a \in U_{1}$ and $f(U_{1}) \subset V$. This is almost what we want, except that we want to find an open set $U$ of $X$ s.t. $U \subset A, a \in U$ and $f(U) \subset V$. This is because $U_{1}$ being open in $A$ does not necessarily mean that it's open in $X$ because $A$ is closed in $X$. (i know how to expand this argument out if needed) My idea now is to try to find a subset $U$ of $U_{1}$ such that $U$ is open in $X$. We have two cases:  (1) $U_{1}$ does not intersect the boundary of $A$. (2) $U_{1}$ intersects the boundary of $A$. Now I'm not sure what to do. I'm wondering if it would help to investigate the relationship between Bd(A) and Bd(B).. In any case (no pun intended), any help/advice would be greatly appreciated. :)  Sincerely,  Vien
15891
Let $f$ denote a map from a product space $X \times Y$ to $Z$. If for every $x\in X$, the map $f(x,-)$ is continuous, and the same holds for every $y \in Y$, then is $f$ continuous in general? If not, is there any condition to be imposed to make $f$ continuous?
19438
The graph of $f$ is $G(f) = \{(x,f(x)) : x\in X\} \subseteq X\times Y$  $X$ and $Y$ are metric spaces.  a) Suppose $f$ is continuous and prove that $G(f)$ is a closed set.  b) Suppose that $G(f)$ is compact and prove that $f$ is continous  For a), the definition of a closed set that comes to my mind is a set that contains all its limit points (or was it accumulation points?), is there another equivalent definition that may b more helpful to prove a)? Is it possible to prove this directly? Because at first glance the only way I could imagine to prove this is by contradiction or contrapositive.   I imagine that the proof of b) will be immediately derived from a).
49003
How to prove: Continuous function maps compact set to compact set using real analysis?  i.e. if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, then $f([a,b])$ is closed and bounded.   I have proved the bounded part. So now I need some insight on how to prove $f([a,b])$ is closed, i.e. $f([a,b])=[c,d]$. From Extreme Value Theorem, we know that $c$ and $d$ can be achieved, but how to prove that if $c &lt; x &lt; d$, then $x \in f([a,b])$ ?  Thanks!
73765
It'd be great if someone checked the proof I did for the following problem:  $f:X\longrightarrow Y$,   $f(\overline{A})\subseteq \overline{f(A)},\forall A\subseteq X \Leftrightarrow f$ continuous  proof:  Suppose $f(\overline{A})\subseteq \overline{f(A)}$ for any $A\subseteq X$ .  Let $C\subseteq Y$ be closed and define $A=f^{-1}(C)$.  $f(\overline{A})\subseteq \overline{f(A)}\subseteq\overline{C}=C$, therefore $\overline{A}\subseteq f^{-1}(C)=A$,  so $A$ is closed.  *Conversely, suppose $f$ is continuous.    Let $p\in\overline{A}$ and let $V$ be an open neighborhood of $f(p)$. Since $f$ is continuous,  $f^{-1}(V)$ is open. On the other hand $p\in f^{-1}(V)$, but since $p\in\overline{A}$ we have that $f^{-1}(V)$ contains points of  $A$. Therefore $V$ contains points of $f(A)$. Since $V$ is an arbitrary neighborhood of $f(p)$, we have $f(p)\in \overline{f(A)}$. So finally, $f(\overline{A})\subseteq\overline{f(A)}$.  Thank you very much!