@Thorgott ok, am finally getting off of work so I’m able to think about this. I see in the case of f=id that I’d be defining h as the constant map to c. So certainly c not in the range is a necessary condition for h to be injective. I guess I’m still stumped in terms of seeing how to argue that f injective and c not in the range get me to the conclusion.
I must somehow be able to show that f^k(c) = f^j(c) means k = j but I still can’t quite see how
@Thorgott OK I think this might get me there but hoping to follow up. Let me consider the corestriction of $f$ to its range and then apply to both sides, so that I get as you say $c=f^{j-k}(c)$ (incidentally, how would you make this precise? induction?). Now we conclude that $j = k$ for otherwise we would have $c \in \ran f$
it might be good self study practice in general, but it severely limits your ability to get useful feedback from third party sources
particularly if it amounts to (as it seems to here) not "how do i do this proof by induction" but "how do i do proof by induction in this particular way"
but, 99% of real life difficulty in induction proofs is figuring out how to structure the argument so that induction is possible (roughly speaking, "figuring out what to induct on"), and i don't know of any textbook that gives a particularly good sense of this
in crummy 'intro to proof' books it is literally always drop dead obvious what you are supposed to induct on, because the problems are basically made up to fit into some recipe
and in something like enderton it's like "wait, do we have the order on natural numbers 'yet' or are we still phrasing everything in terms of the successor operator"
which is certainly a kind of difficulty worth overcoming in induction proofs, but maybe not the most representative kind
and its not well delineated away from the stuff thats like "here is a way to encode ordered pairs as sets" and "here is what an inductive set is"
insofar as it treats the axioms in a depth greater than most textbooks, and consequently teaches more of set theory than any average math PhD would need to know
a lot of people expect a book with 'set theory' or equivalent as its title to be a kind of 'just the facts' handbook, and enderton is more than that
hmm, if you recall specific instances of what you mean i'd definitely be curious. i'm basically done Ch 4 and the "deepest" things I've taken I think relate to proving that certain sets exist
by using axioms like pairing, union, etc to show that some set contains them, and then axiom schema of specification to specify down to that set
I suspect that's because us rookies are drawn to maths given that it promises complete precision, and then you see that you have to go down, down, down... to actually realize that promise. I am content for now going no further than Enderton though
it's not the end of the world to power through anything, and i don't want to make it out like it's scary or inaccessible, i just wonder whether everyone bothers to examine why they want to do set theory
it's like if you were learning a language, at some point in acquiring the rules of grammar, and their big exceptions, and the histories of how words or figures of speech developed from other words or languages, eventually you reach a point where you're learning stuff that > 99.9% of native speakers have never heard of and never use
and with something like set theory, you hit that point sooner than a lot of people think
and yeah i actually wonder how many books are actually worth reading start to finish
a lot of standard textbook advice fits the format "read through chapter x and then stop," where x is often only a little more than halfway through
But surely something to be said for just enjoying a good book :) I agree some are slogs, but some are just really pleasant. (Baby) Rudin fell in the first category (at least for me), but Enderton I am really enjoying. LADR was another beautiful one
LADR was long ago and I'm so out of practice that i gotta do another LA book sadly, hence hoffman and kunze
i believe that some people might have hit chapter 8, because i did, despite xander's advice (which is the best advice and the advice taken by my first analysis prof). but nobody has ever gone through chapters 9, 10, and 11 of rudin. you can't. your brain shuts off.
@Jakobian suppose if there is a sinx term in an expression and the x tends to zero, they multiply and divide by x and take that particular term to be 1
@KavinIshwaran you can't do that unless it makes sense with respect to the rules that limits obey e.g. products, sums, quotients of limits
You would have to be more specific and only then we can discuss if in a given limit it makes sense to make such substitution. In general the answer is no, it doesn't make sense
to be more specific, in a question which goes like this : limit x tends to zero (tanx - xsinx)/x they multiplied sin x by x and divided it with x (that particular term alone and then LH rule is used)
@Jakobian Oh Ok. I will make a memory note on this.
The above is supposed to be an example of a sequence that converges in mean but not pointwise a.e. I simply do not understand it, since I believe there are some typos. How can n increase from $2^m$ to $2^m-1$? Also, is $f_4$ correct? Mightily confused.
Trying to figure out what the correct example should be...
The typewriter sequence is an example of a sequence which converges to zero in measure but does not converge to zero a.e.
Could someone explain why it does not converge to zero a.e.?
$f_n(x) = \mathbb 1_{\left[\frac{n-2^k}{2^k}, \frac{n-2^k+1}{2^k}\right]} \text{, where } 2^k \leqslant n < 2^{k+...
Hi everyone, wondering if someone can point me in the right direction before I resort to asking a question. I'm analysing numerical stability of an operation and can incur in an error that is (number of elements of the vector smaller than machine precision)×ε, which obvs. ≤ (size of the vector)×ε, but this seems super loose?
i hav a 1D manifold which is the real line, R. i define two atlases : Atlas 1 : ψ(x)=x. Atlas 2: ψ(x)=2x,x≥0;ψ(x)=−3x,x<0 why arent these two atlases two different smooth structures?
the first one is a smooth structure becuz it's a homeomorphism onto $R^1$ and there's only one chart so the compatibility of charts holds. the second one is also smooth for the same reason
sorry i meant $\psi (x)=2x ; x\geq0$ and $=3x , x<0$ for the second atlas
@RyderRude what are you objecting. Which part of what I said
@RyderRude there's multiple smooth structure on $\mathbb{R}$ but no matter which one you equip $\mathbb{R}$ with, those will be, up to diffeomorphism, your standard $\mathbb{R}$
I'm reading a proof of the fact that the space $C_c(\mathbb R^n)$ of compactly supported continuous functions on $\mathbb R^n$ is dense in $L^1(\mathbb R^n)$. The proof starts by a reducing the proof to simpler special cases. We can assume $f$ has compact support by DCT and moreover that it's positive. I understand that. However, then it is claimed, since $f$ is measurable, there is an increasing sequence of simple functions of, note, compact support, that converge to $f$.
I understand why there's such a sequence, but I don't understand why the sequence can be compactly supported. Why?
@YourLordJoyBoy Tags can be given descriptions. You generally shouldn't touch that since if you add a useless tag and give it a description then it will stay forever until it gets moderated
Unless you're trying to give description to an already existing tag
> Theorem 2.25. A subset $A \subset \mathbb{R}^{n}$ is Lebesgue measurable if and only if for every $\epsilon>0$ there is an open set $G$ and a closed set $F$ such that $G \supset A \supset F$ and $$\mu(G \backslash F)<\epsilon.$$ If $\mu(A)<\infty$, then $F$ may be chosen to be compact.
I'm reading a proof of the fact that the space $C_c(\mathbb R^n)$ of compactly supported continuous functions on $\mathbb R^n$ is dense in $L^1(\mathbb R^n)$. Let $A$ be a bounded set. Somewhat out of the blue, it is claimed that from the Borel regularity of Lebesgue measure, there exists a bounded open set $G$ and a compact set $F$ such that $G \supset A \supset F$ and $\mu(G \setminus F)<\epsilon$.
I know the theorem above, but it doesn't say that we can choose $G$ to be bounded. Does anyone have any clue why $G$ can be bounded?
the simpler question might be, what is the quickest path in your document to seeing that G can be bounded. i might look at the proof of "theorem 2.25" to see where the existence of its "G" comes from. it might be as simple as tracing through that argument
as a background vibe, you could think about covering of A by "simpler" open sets (e.g. dyadic rectangles, or other sets whose diameter you can control), and convince yourself that if A is bounded and the sets in your cover have controlled enough diameters, the union of the cover will be bounded too
but that might not be the simplest route in your reference
I think I found an answer. If $A\subset\mathbb R^n$, then $\mu^\ast(A)=\inf\{\mu(G):A\subset G, G \text{ open }\}$. If $A$ is measurable, we can find an open set $G$ such that $\mu(G)<\mu(A)+\epsilon$ for some $\epsilon>0$, and if $\mu(A)<\infty$, we'll have $\mu(G)<\infty$.
@AlessandroCodenotti I figured there exist spaces which are Dieudonne complete but not Cech complete, and there are spaces which are Cech complete but not Dieudonne complete
