Let $C$ be closed set in $\Bbb R$ and let $f:C\to \Bbb R$ be continuous.  Show that there is a continuous function $g:\Bbb R\to \Bbb R$ with $g(x)=f(x)$ for every $x\in C$.  I am unable to view how to construct $g$ from $f$.  Please give some hints.
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 $E$?
87911
Let $f: \Bbb R \to \Bbb R$ be a continuous function satisfying $f(x+n) \to \infty$ as a sequence in $n$, for all $x$. Does $f$ satisfy $f(x) \to \infty$ as $x\to \infty$?   If we drop the continuity assumption then the claim is false, by considering a function tending more and more slowly to $\infty$ as we start at larger values in $(0,1)$. (Or many other examples)  As for context: a variant of this claim (when $f$ is analytic and we replace $n$ by a general increasing sequence $a_n$) could be useful to me at some technical exercise, and this is a simplification which I still cannot tackle.  Assuming by contradiction that $\exists M\forall x \exists x_0&gt;x: f(x_0)\leqslant M$, we want to show that $\exists x \exists N&gt;0 \forall n \exists n_0&gt;n: f(x+n) \leqslant N$. No "quantifiers-level" logic seems to apply here.  Any ideas?
64888
If $f,g$ are continuous at $a$, show that $h(x)=\max\{f(x),g(x)\}$ and $k(x)=\min\{f(x),g(x)\}$ are also continuous at $a$. Here is my attempt at a proof. It feels very elaborate and I am not sure if it is correct. Can someone please point out any mistakes or places where I may improve. Thanks!  By the definition of continuity at $a$ of $f,g$ we have that $\lim\limits_{x\to a}f(x)=f(a)$ and $\lim\limits_{x\to a}g(x)=g(a)$. Suppose that $f(a)&gt;g(a)$. Then there exists a $\delta&gt;0$ such that $f&gt;g$ for all $x$ satisfying $|x-a|&lt;\delta$. Then for $x$ satisfying $|x-a|&lt;\delta$ we have $h(x)=f(x)$ and $k(x)=g(x)$ and thus $h$ and $k$ are continuous at $a$ because $f$ and $g$ are continuous at $a$. Now if $f(a)&lt;g(a)$ we simply relabel $f=\tilde{g}$ and $g=\tilde{f}$, so that $\tilde{f}(a)&gt;\tilde{g}(a)$ and we apply the previous result to show that again $h$ and $k$ are continuous at $a$.\  Now if $f(a)=g(a)$ then we can distinguish three cases. $f\geq g$ in a small neighborhood about $a$, $f\leq g$  in a small neighborhood about $a$ or $f\geq g$ on one side of $a$ and $f\leq g$ on the other side. In the first case we assume that $f\geq g$ in some small neighborhood about $a$. Then, since $f(a)=g(a)$ we have $h(x) = f(x)$ in this neighborhood and $k(x) = g(x)$ and again we see that $h$ and $k$ are continuous at $a$ due to the continuity of $f$ and $g$ at $a$. Similarly if $f\leq g$ in a neighborhood around $a$ then $h(x)=g(x)$ and $k(x)=f(x)$ in this neighborhood and $h$ and $k$ are continuous at $a$.   Suppose that just to the left of $a$ we have $f\geq g$ and just to the right of $a$ we have $f\leq g$. Then $\lim\limits_{x\to a^-}h(x)=\lim\limits_{x\to a^-}f(x)=f(a)=h(a)=g(a)=\lim\limits_{x\to a^+}g(x)=\lim\limits_{x\to a^+}h(x)$. Similarly $\lim\limits_{x\to a^-}k(x)=\lim\limits_{x\to a^-}g(x)=g(a)=k(a)=f(a)=\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a^+}k(x)$. So $\lim\limits_{x\to a}h(x)=h(a)$ and $\lim\limits_{x\to a}k(x)=k(a)$ which means that $h$ and $k$ are continuous at $a$. Lastly if $f\leq g$ just to the left of $a$ and $f\geq g$ just to the right of $a$ we can relabel $f=\tilde{g}$ and $g=\tilde{f}$ and apply the last result to show $h$ and $k$ are continuous at $a$.   Again, please point out any mistakes I may have made. Thanks!!
97034
I'm trying to do the following:     Prove that a function $f : S\rightarrow T$ between two topological spaces is continuous iff $f^{-1}(C)$ is closed whenever $C\subset T$ is closed.   To show that the closed-set relationship implies continuity, notice that $T-C$ is open. The preimage of this open set is $f^{-1}\left(T-C\right)=S-f^{-1}(C)$, which is open by assumption ($f^{-1}(C)$ is closed). Of course, any open set $G\subset T$ can be written in the form $T-C$ by letting $C=T-G$. Thus, for every open set $G\subset T$, the preimage is open.  To show the implication in the other way, consider a closed set $C\subset T$. $T-C$ is open, so $f^{-1}(T-C)=S-f^{-1}(C)$ is open as well ($f$ is continuous), which means that $f^{-1}(C)$ is closed.
17962
Let $\,f:[a,b] \rightarrow  \Bbb R $ be  continuous  and  $\int_a^b f(x)g(x)\,dx=0$, whenever $g:[a,b] \rightarrow  \Bbb R $ is  continuous and $\int_a^b g(x)\,dx=0$.  Show that $f$ is a constant function.  I tried a bunch of things including the mid-point integral theorem(?) but to no avail. I'd appreciate an explanation of a solution because I really don't see where to go with this one..
38425
My question: let $f : X \to \Bbb R$ be a non-constant continuous function on a connected metric space and assume that $f(X)$ is uncountable; then $X$ is uncountable.  We know continuous image of a connected metric space is connected. Since connected space in $\Bbb R$ are intervals and the function is non-constant continuous function so $f(X)$ is an interval in $\Bbb R$ and hence uncountable.  Is the proof correct?? How does it follows that $X$ is uncountable?
12516
Let $f: X \to Y$ be a function. The graph of $f$ is defined to be the set $G = \{(x, f(x)) : x \in X\}$. Prove that if $f$ is continuous then $G$ is homeomorphic to $X$.     an anyone suggest  me the process of solving this problem, thanks.
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!
49805
As the title indicates, I'd like to prove the following:     If $f:\mathbb R\to\mathbb R$ is a continuous function on $[a,b]$, then $f$ attains its maximum.   Now, I do have a working proof: $[a,b]$ is a connected, compact space, which means that because $f$ is continuous, $f([a,b])$ is compact and connected as well.  Therefore, $f([a,b])$ is a closed interval, which means it has both a minimum and, as desired, a maximum.  What I would like, however, is a proof that doesn't require such general or sophisticated framework.  In particular, I'd like to know if there's a proof that is understandable to somebody beginning calculus, one that (at the very least) doesn't invoke compactness.  Any comments, hints, or solutions are welcome and apreciated.
26869
If $|f(x)|$ is continuous at $a$, is $f(x)$ continuous at $a$?  I tried doing it using composite functions. If $g(x)= |x|$, then $g\circ f(x)= |f(x)|$. Since $g(x)$ and $g\circ f(x)$ are continuous, $f(x)$ is continuous.  I don't know if this is correct. Please help.
23034
Let $f : (X, d_X) \to (Y, d_Y )$ be a map between two metric spaces. Recall that $f$ is called continuous at $x ∈ X$ if for any open ball $ B_f(x)(ϵ)$ of radius $ϵ$ around $f(x)$ there exists a ball $B_x(δ)$ of radius $δ$ around $x$ such that $f(B_x(δ))⊂B_f(x)(ϵ)$. Show that $f$ is continuous at $x$ if and only if for any sequence ${x_n}$ in $X$ converging to $x$ the sequence $f(x_n)$ converges in $Y$ to $f(x)$. Hint: imitate the proof of the corresponding statement for functions $R → R$.
23644
I'm new here. Can anyone give me a hint in this problem? Please.  Let $X$ be a topological space. If $f,g$ are continuous function of $f,g\colon X\to S^n$  which $f(x)\neq -g(x)$ for all $x$ in $X$, show $f$ and $g$ are homotopics. I try to show a homotopy using $$q(x) = \frac{f(x)+g(x)}{\|f(x)+g(x)\|}$$ it's well defined, but i'm stuck.
46530
Suppose $f$ and $g$ are continuous on $[a,b]$ and that $f(a) &lt; g(a)$ but $f(b) &gt; g(b)$.   Prove that $f(x) = g(x)$ for some $x \in [a,b]$.     I did try and actually did get as far as coming up with $f-g$. Would I set $h(x)= f(x)- g(x)$ and say $h(a) &lt;x&lt; h(b)$?
5443
Let $X$ be a metric space with metric $d$. Define $d: X \times X \to \mathbb{R}$, show that $d$ is continuous.  I would like to show that the function is continuous the topology way (since it is a course on topology).   Let $(a,b)$ be a basic open set in $\mathbb{R}$, then $d^{-1}(a,b) = \{(x,y): a&lt;d(x,y)&lt;b\}$. Define $A= \bigcup_{x \in X} B_{b} (x)$, where $B_b (x) = \{y \in X| d(x,y)&lt;b$}. Clearly $A$ is open. Now define $C = \{(x,y):d(x,y)&gt;a \}.$ Essentially I hope that $A \cap C$ is open. But I am stuck in showing that $C$ is open.  Any help please? :)
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?
16363
Hello I am doing this exercise where we let X and Y be topological spaces and $f:X\to Y$ a function. We suppose that $\mathcal{A}$ is a collection of closed subsets of X, such that $\bigcup_{A\in\mathcal{A}} A = X$. We assume that for every $A\in \mathcal{A}$ the restriction $f_{|A}: A\to Y$ of f to A is continuous. Furthermore we suppose that for each $x\in X$ there exists a neighborhood U of x such that the intersection $A\cap U$ is non-empty for only finitely many $A\in\mathcal{A}$. And I have to show that f is continuous. I know that since $f_{|A}: A\to Y$ is continuous we have that for all open subsets $O\subseteq Y$ the inverse image $f^{-1}(O)$ is an open subset of A. And to show that $f: X\to Y$ is continuous I am thinking of showing that for each open subsets $O\subseteq Y$ the inverse image $f^{-1}(O)$ is an open subset of X, but this would be obvious wouldn't it since for all open $O\subseteq Y$ we have $f^{-1}(O)\subseteq A\subseteq X$?
42554
Find all the continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that if $x-y\in \mathbb{Q}$ then $f(x)-f(y)\in \mathbb{Q}$.   Can somebody give me some hint to solve this problem. I think that the functions are $f(x)=mx+b$ where $m\in\mathbb{Q}$.  Thanks!