@MattN okay, point taken. But in fact there's a very quick proof of Urysohn's lemma for metric spaces: given $A$ and $B$ disjoint, nonempty and closed, put $f(x) = \frac{d(x,A)}{d(x,B) + d(x,A)}$ to get an Urysohn function.
That's much simpler than the onion shell argument.
Tietze for metric spaces: Let $A \subset X$ be a closed subset of a metric space and let $f: A \to [0,1]$ be a continuous function. Hausdorff has given the following formula for a Tietze extension $F$ of $f$: $$F(x) = \begin{cases} f(x), & \text{if } x \in A, \\ \inf{\left\{f(a) + \frac{d(x,a)}{d(x,A)} - 1\,:\,a \in A\right\}}, & \text{if } x \in X \smallsetminus A.\end{cases}$$ It is not hard to verify that $F$ is indeed continuous.
@MattN Of course you can. If $f(y) = f'(y)$ then add something to $f'$ that is non-zero on $y$ (e.g. $d(\cdot, g(X))$ if $y \in Y \smallsetminus g(X)$).
The problem is that you don't say what functions you want, and you want that $f=f'$ on $g(X)$ to get a contradiction to injectivity.
Yes. You could take any old function that is zero on $g(X)$ and $1$ on $y \in Y \smallsetminus g(X)$ and add it to any function on $Y$ to get an $f'$.
How can I express $$\bar{F}=(x^2+y^2+z^2)(x\bar{i}+y\bar{j}+z\bar{k})$$ in polar coordinates? I have seen elementary cartesian-to-polar -conversions with $x=R\cos(\theta)$ and $y=R\sin(\theta)$ but how is it here?
@tb: how can I calculate nabla after the coordinates conversion? In cartesian, I had $\partial_{x}, \partial_{y}, \partial_{z}$ but now I have only $R, \theta, \alpha$ but $R$ is constant?!
@Nunoxic I find one file per chapter slightly easier to work with -- but that depends a lot of which kind of editor you're using. Some editors make it easier to work with many files than to work at many simultaneous points in a single file; others are the opposite.
@Nunoxic I think is better to split your work in one file per chapter. You can create a master document with the preamble and all the global options that you want for your document, and you can call each chapter by using command \include{}
@hhh you are missing a right brace and a \left before a left paren: $\left(\hat{e}_{R}\partial_{R}+\frac{1}{R}\hat{e}_{\theta}\partial_{\theta}+ \frac{1}{R}\sin(\theta) \hat{e}_{\phi}\partial_{\phi}\right)\cdot\bar{F}$
@Nunoxic Texmaker have a option to work in "Master Document" mode, so that if are working in a chapter and compile, Texmaker compile the master document
@robjohn The problem is that things such as $\hat{i}$ do not appear on the left-hand-side and I should do dot-product?! It will give zero?! Cannot, there must be something wrong -- ideas?
@Nunoxic you can select only the chapters that you want to compile by using \includeonly{} command. His mandatory must be a comma-separated list of names of files that you want to compile
@robjohn Is this $\hat{i}=\frac{\partial_{x}\bar{F}}{|\partial_{x}\bar{F}|}$ right? I need somehow to convert the $\hat{i}, \hat{j},\hat{k}$ in terms of something else, thinking...
@robjohn I am not sure what this infers -- so I can just multiple first rows $\hat{i}$ with $\hat{e}_{\theta}$? (without caring about specific basis...?! No cannot be like that...they must be of the same type?! Second, checking calcs...)
@robjohn How did you deduce that? You divide each coordinate by the length of $R$?! One must be able to deduce that with partial derivatives somehow, thinking...
@MattN there is ambiguity in language, I guess. Some people reserve "isometry" for a surjective isometric map while others don't. I would think that in view of 2. and 3. the surjective case looks more reasonable (else what's the difference between 3. and 4.?) Also, in 3. 2. of the new version, you want $f''|_{g(X)} = 0$ otherwise you don't get $T(f) = T(f')$.
At the very beginning of 1. you could add a word why $\|f \circ g\| \leq \|f\|$.
I skipped the background and just read the "possible cases: Vitali set is finite; countably infinite; has at least $\aleph_1$ many elements", then wrote him an answer about those cases.
and then you have to multiply both sides of the equation by that integrating factor and integrate both sides. I think this is where the tactic comes in handy because on the left your just left with y*x^3=int(500x^6ln(x))
or sorry i meant to say the integral of x^3(500x^6 ln(x)
then u integrate that leaving the C behind on the right and then solve for it using the conditions present in the problem
@user979616 I was looking up through the transcript to find the original question, so that I might help you. I failed, but if you want to post again, I'll try to help you.
@Skullpatrol - "from down under" is OK - this includes Australia and New Zealand. However "the land down under" usually just means Australia. New Zealanders get offended when people mistake them for Australians.
@MattN I find what you write rather confusing. You should first argue why $\|T(f)\| \leq \|f\|$ (you just claim that so far) and then argue that the constant functions show that $\|T\| = 1$
This (together with continuity) also shows that equality $\|T(f)\| = \|f\|$ holds for all $f$ if and only if $g(X)$ is dense (hence all of $Y$ by compactness).
Yes, it is a good exercise. By the way: you're close to ten edits (which turns your answer automatically to Community Wiki mode), so either post a new answer or flag for mod attention to undo the CW mode if you want to earn some points for the hard work.
Related to page 819 prob 4 in this book. I calculated wrongly:
$$(\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}) \cdot (\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}) =\partial_{r}\partial_{r}+\frac{1}{r^2}\partial_{\theta}\parti...
I was thinking about our conversation yesterday, about the phrase "what a cute baby gigili migili", or whatever it was. I remembered that my son, at a very young age, coined the word gigili in an extremely similar context. I had completely forgotten until this morning.
So there's this baby tickling thing that they do in former Yugoslavia. You walk your fingers up the baby's arm and tickle their armpit. But the accompanying words are "Ide bubamara, nema gdje da spava, traži krevetić. Gili gili gić."
Related to page 819 prob 4 in this book. I calculated wrongly:
$$(\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}) \cdot (\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}) =\partial_{r}\partial_{r}+\frac{1}{r^2}\partial_{\theta}\parti...
