Yes, I was talking about ##Y##. But both the approaches given use a shrinking family of neighborhoods, which seems to suggest at least that we deal with first countable spaces?
Ah nice, a more topological approach. I didn't write out the details of your post but this seems to suggest that the statement is more generally true than in metric spaces. I would be interested in knowing what minimal assumptions are such that this holds. I guess the Hausdorffness condition and...
@etotheipi is not wrong. Linearity refers (in the context of abstract algebra) to both the preservation of the scalar multiplication and the addition. A linear map is a map that is both homogeneous and additive.
Close enough. The reverse inequality is not quite the definition of continuity but you need a routine triangle inequality argument to get there, like you did in the other inclusion. But I'm sure you could have filled up the little gap. Well done! I think your solution is the the shortest possible.
Indeed for this first set equality you needed something like injectivity which is not given.
Can you edit or make a new attempt at the question so everything reads smoothly? Otherwise everything is shattered through multiple posts making it very hard to read for others (and me when I go through...
I think there is a problem with your intersection, at least semantically. You have an intersection ##\bigcap_{n \in \Bbb{N}} A_n## with ##A_n \subseteq \mathcal{P}(X)##. Thus semantically your intersection is a subset of ##\mathcal{P}(X)##. However, the left hand side is ##\{x\}## which is a...
Question 1 and question 2 are dual (use complements), so I will only answer question 1.
Q1: Axler claims that the set ##\mathcal{B}:=\{\bigcap \epsilon\mid \epsilon \mathrm{\ countable \ collection \ of \ opens}\}## is not equal to all Borel sets. His argument is: showing that there exists a...
I'm not sure what you are after:
$$\frac{|f(x)-f(x_0)|}{5} = \frac{5|x-x_0|}{5}= |x-x_0|$$
and the inequality you are looking for so hard is always true (it is even equality!) regardless of the fact that ##|x-x_0|## is small.
I agree with @PeroK that the author was probably sloppy and yes it...
I'm very sorry. I made a mistake in my definition of ##\mathcal{S}## in post #2 (now fixed), which is probably why your attempt doesn't work because my definition of ##\mathcal{S}## contained a mistake. I guess that happens when I give hints without writing down anything on paper ;)
But as you...
Your first solution + the clarification in post #54 solves the question (I didn't look at the second one though)! Well done! I guess I must come up with less routine exercises since you seem to solve all of them ;)
This is much more readable to me than your previous post. Your approach has all the right ideas. Especially the line "The only continuous functions of the form ##2\pi n(\theta)## are constant functions." is the key to an elementary approach that does not use black magic. I consider this...
You should show an attempt.
Can you at least show that ##\mathcal{S}:=\{E \subseteq X: E \mathrm{\ countable \ or \ E^c \ countable}\}## is a ##\sigma##-algebra containing ##\mathcal{A}##? Why is it the smallest?
Axler does not say that. The collection of all subsets is a ##\sigma##-algebra (trivially). Axler says that we cannot define Lebesgue-measure on this ##\sigma##-algebra and that's why we define Lebesgue measure on Borel sets.
But if the students can solve it correctly, so should the teacher! I'm speaking high school here. At the university level I can imagine this happening.
You really need that ##g(\Bbb{C}^*) = \Bbb{C}^*## for your argument to work. Also, when using the symbol ##\cong##, explain what you mean with the symbol. I assume that you mean isomorphism in the category of continuous maps, but how are you even sure the inverse on the image is continuous as well?
Seems to work! Working with diagonal operators was also what I had in mind. Here is a more sophisticated approach using the theory of ##C^*##-algebras:
Consider the ##C^*##-algebra ##C(K)##. Then the inclusion ##i: K \to \Bbb{C}## has spectrum ##K##. Next, choose a Hilbert space ##H## and an...
But the plane ##\Bbb{C}## with the origin removed is not simply connected (consider fundamental group), so the theorem does not apply. Also, if the plane was simply connected your theorem would contradict the exercise.
There are multiple approaches. First, you need to say what convention you use: I guess you write ##z=r e^ {i \theta}##. What domain do you allow for ##\theta##?
Then, why does ##g## take that form and why does it have a discontinuity?