Here is the full question: Let $X$ be a topological space and $f,g$ two continuous functions $f:X\to \Bbb R$. Prove that the functions $h(x)=|f(x)|$ and $k(x)=\max\{f(x),g(x)\}$ are both continuous. I am having trouble proving the first part that $h(x)=|f(x)|$ is continuous. Here's my answer for the second part: We know the maximum of two functions $f(x)$ and $g(x)$ is defined as $max\{f(x),g(x)\}=\frac{f(x)+g(x)}{2}+\frac{|f(x)-g(x)|}{2}$. Since this is the sum of two continuous functions $f(x)$ and $g(x)$, then by the properties of continuous functions, $k(x)$ is also a continuous function. Could someone point me in the right direction for the first part?
35514
Let $f(x)$ be a continuous function. Prove that $\left|f(x)\right|$ is also continuous.  Is it correct to say that, by the reverse triangle inequality, $\left|f(x)-f(c)\right| \geq \left|f(x)\right|-\left|f(c)\right|$ in all cases, so we will always have that for all $\left|x-c\right|&lt;\delta$ implies $\left|f(x)-f(c)\right|\leq \epsilon$?  I am not sure if my solution adequately uses the functional form, since I just used the equation and not my actual function. Please help with the proper solution!
110530
Suppose $f:\Bbb R\to\Bbb R$ is continuous such that $f(x)\to0$ as $|x|\to\infty$. Prove that $f$ attains either a maximum or a minimum.   My attempt at the question :  Given $\epsilon &gt; 0 \ \ \exists \ \alpha &gt; 0$ such that $\ |f(x)| &lt; \epsilon\ \ \ \forall \ \ \  \ |x| &gt; \alpha$    Now, when  $\ \ x \in [-\alpha,\ \alpha] $, by the Extreme value theorem, since $f(x)$ is a continuous function on a closed interval, it attains a maximum (say $ M$) and a minimum (say $m$).    I am stuck at this part. I think the way to move forward is by using the fact that we can make $ \epsilon $ as small as possible. But I don't know how to prove that it will surely attain a maximum or a minimum or both.
10561
In a  I needed an argument of the following kind:  Let $f, g: [a, b] \to \Bbb R$ be functions with the following properties:  $f$ is differentiable, $g$ is continuous, $|f'(x)| \le g(x)$ for all $x \in [a, b]$.  Then $|f(b) - f(a)| \le \int_a^b g(x) \, dx$.  Seems pretty easy: We have $$ \tag{*}  |f(b) - f(a)| = \left| \int_a^b f'(t) \, dt \right|  \le \int_a^b |f'(x)| \, dx \le \int_a^b g(x) \, dx \, . $$ There is just one problem: We have used the , and that requires $f'$ to be Riemann integrable (or Lebesgue integrable and absolutely continuous), compare . This is for example satisfied if $f'$ is continuous. But the above statement holds without additional assumptions on $f'$: Let $n$ be a positive integer and $x_k = a + \frac kn (b-a)$, $0 \le k \le n$, be a partition of the interval $[a, b]$. We apply the  to each subinterval $[x_k, x_{k+1}]$: $$  f(x_{k+1}) - f(x_k) = (x_{k+1} - x_k) f'(c_{n, k}) $$ for some $c_{n, k} \in [x_k, x_{k+1}]$. It follows that $$ |f(b) - f(a)| \le \sum_{k=0}^{n-1}| f(x_{k+1}) - f(x_k)| \le \sum_{k=0}^{n-1} (x_{k+1} - x_k) g(c_{n, k}) \, . $$ $g$ is continuous and therefore Riemann integrable. The expression on the right is a Riemann sum for $\int_a^b g(x) \, dx$ with the partition size $(b-a)/n$. It follows that $$  \lim_{n \to \infty } \sum_{k=0}^{n-1} (x_{k+1} - x_k) g(c_{n, k})  = \int_a^b g(x) \, dx $$ and therefore $|f(b) - f(a)| \le \int_a^b g(x) \, dx$. My question: Is there a simpler proof of the statement? It seems pretty basic, therefore I wonder if there is a simpler proof, without going into technical details such as partitions and Riemann sums. Perhaps it is a consequence of some other theorem about integration which I failed to find?
47071
Let $f:\mathbb{R}\to\mathbb{R}$ be a map sending closed intervals to closed intervals. Prove that $f$ is continuous or find a counter example.  WLOG we just have to prove continuity at $0$ and we can also assume that $f(0)=0$ and $f(1)=1$ (by considering $x\mapsto (f(x)-f(0))/(f(1)-f(0))$ ).  I build up this question while thinking of the question .
41991
Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a continuous real function and we consider its graph identified by the subset:   $$\mathcal{G}(f)=\{(x,f(x))\in\mathbb{R}^{n+1}:x\in\mathbb{R}^n\}$$  of $\mathbb{R}^{n+1}$ equipped with the usual euclidean topology that induced on $\mathcal{G}(f)$ a subspace topology. Well with these condictions we demonstrate that $\mathcal{G}(f)$ is homeomorphic to $\mathbb{R}^n$. So we consider the function   $$ h:\mathcal{G}(f)\owns(x,f(x))\rightarrow x\in \mathbb{R}^n$$  since the domain of $f$ is $\mathbb{R}^n$, clearly $h$ is surjective on $\mathbb{R}^n$ and then is also injective because if $h(x,f(x))=h(y,f(y))$ then $x=y$; now to prove the assertion we have to demonstrate of that $h$ and $h^{-1}$ are continuous funcion or that $h$ is open and continuous, but unfortunately I'm not be able to do this so can someone help me?
119864
I have some exercise that asks me this:  Find $f$ and $g$ discontinuos such that $f+g$ is continuous. This is what I tought:  $$f = \mbox{sign}(x)$$ $$g = -\mbox{sign}(x)$$  Where $\mbox{sign}(x)$ is the function that maps to $1$ if $x\ge0$ and $-1$ if $x&lt;0$. Then, the two are discontinuous in $0$, but the sum, is continuous, because it's equal to $0$ in every point. First of all, is this example correct?  The exercise also asks me to find examples such that:  $f,g$ discontinuous, but $f$ composed with $g$ is continuous. (need help) $f$ continuous, $g$ discontinuous but the composite $f$ with $g$ is continuous.   (for the second one, if I take $g=\mbox{sign}(x)$ and $f=|x|$, then $fog = 1$. Is this right?  Could you guys give at least a hint, or a less poor example?
16746
Let $(M,d)$ be a compact metric space and $d(f(x),f(y)) &lt; d(x,y) $ for all $ x\neq y$ Prove that if $f$ is a continuous fuction then there is a unique $x_0 \in M $ such that $f(x_0)=x_0$ I know that since $g(x) = d((f(x),f(y))$ is continuous then a continuous function in a compact space reaches its maximum and minimum. Then by proving that the minimum of $g(x)$ = 0 can I conclude that $x_0$ is a fixed point?
46933
I tried to solve this problem but I could not continue  Let $f: [a, b​​]\to \mathbb{R}$ be a continuous function.  Prove that if $x_1, x_2,\ldots,x_n  \in (a,b​)$, then exists $x_0 \in (a, b​​)$ such that $$f (x_0) =\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}$$  the function is continuous, so $\forall\ x_1, x_2,\ldots,x_n  \in(a, b​​)$ $$\begin{align*} \lim_{x \to \ x_1}f(x)&amp;=f (x_1)=L_1\\ \lim_{x \to \ x_2}f(x)&amp;=f (x_2)=L_2\\ \lim_{x \to \ x_3}f(x)&amp;=f (x_3)=L_3\\ &amp;\vdots\\ \lim_{x \to \ x_n}f(x)&amp;=f (x_n)=L_n\\ \end{align*}$$  We have that:  $$\lim_{x \to \ x_0}f(x)=f (x_0)=\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}= \frac{1}{n}\sum_{j=1}^n f(x_j)$$ Now $$\min\left \{ L_1,L_2,L_3,\cdots,L_n \right \}\le\frac{1}{n}\sum_{j=1}^n f(x_j)\le\max\left \{ L_1,L_2,L_3,\cdots,L_n \right \}$$  and at this point I'm lost ...
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.
119043
Let $f$, $g$, $h \colon \mathbf{R} \to \mathbf{R}^\omega$ be defined by   $$\begin{align*} f(t)&amp;:=(t,2t,3t,\ldots),\\\\ g(t)&amp;:=(t,t,t,\ldots),\\\\ h(t)&amp;:=\left(t,\frac{t}{2},\frac{t}{3},\ldots\right) \end{align*}$$  for all $t \in \mathbf{R}$. How to determine whether these functions are continuous relative to the uniform topology on $\mathbf{R}^\omega$? We of course assume that $\mathbf{R}$ is given the usual topology.
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!
44837
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function.  Show that the points at which $f$ is continuous is a $G_{\delta}$ set.  $$A_n = \{ x \in \mathbb{R} | x \in B(x,r) \text{ open }, f(x'')-f(x')&lt;\frac{1}{n}, \forall x',x'' \in B(x)\}$$  I saw that this proof was already on here, but I wanted to confirm and flesh out more details.  "$\Rightarrow$" If f is continuous at $x$, then $f(x'')-f(x')&lt;\frac{1}{n}$ for $x'',x' \in B(x, r_{n})$.  That is, there is a ball of radius $r$ where $r$ depends on $n$.  Then $x \in A_n$ and thus $x \in \cap A_n$.  "$\Leftarrow$" If $x \in \cap A_n$, then there is an $\epsilon &gt; 0$ and a $\delta &gt; 0$ such that $x' , x'' \in B(x, \delta_n)$ for all $n$ and $$|f(x'')-f(x')|&lt;\epsilon.$$ Take $\epsilon = \frac{1}{n}$.
99142
Let $X$ and $Y$ be two topological spaces.   Prove that for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$ , if and only if $f: X \to Y$ is continuous.   I am stuck on the converse. Suppose $f: X \to Y$ is continuous. Then for every closed set C in Y, $f^{-1}(C)$ is closed in X.   WTS  for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$.     Since $\overline{f(A)}$ is a closed set in Y, $f^{-1} \circ \overline{f(A)}$ = $ \overline{ f^{-1} \circ \overline{f(A)} }$  Also, $f(\overline{A}) \subseteq \overline{ f(\overline{A})}$
69524
Let $X=C[0,1]$ be the set of all continuous functions on $[0,1]$. For any two functions $f,g\in X$, set $$d(f,g)=\|f-g\| = \sup \limits_{0\leq x \leq 1} |f(x)-g(x)|.$$  Prove $(X,d)$ is a metric space.    My attempt:  This question comes down to showing that $d$ is a metric, which means that:   $d(f,g)\geq 0$ $d(f,g)=0\iff f=g$ $d(f,g)=d(g,f)$ Triangular sadness.     1) Nonnegativity is guaranteed by taking the supremum of the absolute value of a sum, since the absolute value of a sum is non-negative, the supremum of non-negative numbers is non-negative.  $\square$    2) For the supremum of $|f(x)-g(x)|$ to be equal to $0$, the greatest number for $|f(x)-g(x)|$ from $0\leq x\leq 1$ must be $0$, hence $|f(x)-g(x)|=0$ for all values on this range, and thus $f(x)=g(x)$ is the constant zero map.  $\square$    3) $|f(x)-g(x)|=|g(x)-f(x)|$ implies that $\sup \limits_{0\leq x \leq 1} |f(x)-g(x)|=\sup \limits_{0\leq x \leq 1} |g(x)-f(x)|$ $\square$    4) Triangular sadness:  We need to prove that $d(f,g)\leq d(f,h) + d(g,h)$  I.e.  $\sup \limits_{0\leq x \leq 1} |f(x)-g(x)|\leq \sup \limits_{0\leq x \leq 1} |f(x)-h(x)|+\sup \limits_{0\leq x \leq 1} |g(x)-h(x)|$  Now I am not sure how I can 'behave' with the sup 'function'. Can I do the following?  $$\sup|f(x)-g(x)| = \sup|f(x)-g(x)+h(x)-h(x)|=\sup|f(x)-h(x)+(h(x)-g(x))|$$ Then use normal triangular inequality:  $$\sup|f(x)-h(x)+(h(x)-g(x))|\leq \sup(|f(x)-h(x)|+|h(x)-g(x)|)=\sup(|f(x)-h(x)|+|g(x)-h(x)|)$$   This is obviously less than $\sup \limits_{0\leq x \leq 1} |f(x)-h(x)|+\sup \limits_{0\leq x \leq 1} |g(x)-h(x)|$, but I don't know how to prove it. How do I finish that one off.
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!!
23132
Assume $f$ is a function over $\mathbb{R}$ satisfying $f(x+y) = f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Show that if $f$ is continuous, then $f(x) = cx$ for all $x \in \mathbb{R}.$   I find it hard to use the definition of continuity to derive anything from the equation. I think we could write the definition of the limit in terms to get $$\forall \epsilon,\exists \delta \quad 0&lt;|x-a|&lt;\delta \quad \implies \quad |f(x)-L| &lt; \epsilon.$$ I am not sure where to go from here.
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!