Because if we exclude U in \tau, it is not a topology due to that fact that you just excluded the open subsets which is basically want a topology is (the collection of open sets.
I couldn't find them
If you can't help then I will ask someone else if you would prefer.
I did solve a cool topology problem for HW last night: Let $X$ be an ordered set with the order topology, and $Y$ be a subset of $X$ under the subspace topology. Find an example of such an $X$ and $Y$ where the subspace topology on $Y$ cannot be realized as the order topology on $Y$ for any ordering.
The difference is that some authors use "countable" to mean infinite with cardinality equal to $\mathbb{N}$ whereas other authors use "countable" to mean those sets together with finite sets.
ok thats what I thought you were doing, its wrong. Try again, and use the actual root symbols for clarity. Remember, the only way something comes our of a square root twice is if its under the root four times
or, if you have two separaye roots, one time under each root
For the union stipulation, show that if $\{U_i\}$ are open, then $\displaystyle \bigcup_i U_i$ is open by showing that $X \setminus \bigcup_i U_i$ is countable.
I've closed the book already, but I think Example 3 gives you some tools to use.
The finite intersection stipulation is easy, fortunately. If $\{U_i\}$ is a finite collection of opens, then it's somewhat obvious that $X \setminus \bigcap_i U_i$ is open.
I.e. $\displaystyle \bigcap_i U_i \subset U_k$ for any $k$.
Oh. And by assuming a {U_i} to be countabley infinite of opens. , We can say that the complement X-U is also infinitely countable of opens. Thus Proving our first point
Sadly I won't be able to even go onto the next chapter until I do all the problems and I think that I have mastered the chap. And I have one rigorous proof and 5 other problems!
Basically whenever you are given a topology (a collection of sets which we call open), one of the things you want to check is that the topology is indeed closed under infinite unions.
It's not a theorem, but rather a stipulation that needs to be checked if we wish to determine if something is indeed a topology.
OK. Yes it is the same problem. I am assuming the collection {U_i} to be countabley infinite of opens. thus making X-U as well. Then I want to prove that the arbitrary unions and finite intersections of X-U are closed ?
I'm working on my homework and being asked to prove "If $A\subseteq B\subseteq C$ and $A$ and $C$ are countably infinite, then B is countably infinite." Right now I'm just trying to understand the problem. Would $B$ also be countably infinite if we only had $B\subseteq C$ and $B\neq\varnothing$ ?
I think I have a fair enough grasp on the direction of the proof, but I wasn't planning on making particular use of $A$, so I was left uncertain of its significance
@ABeautifulMind Related to the starred comment, I think that pain the video is referring to comes from very hard work and many sacrifices one has to do.
Without hard work, but wait, I think things are possibly not clearly understood, it's about extremely hard work, and without it I don't think I would have ever imagined to write books, articles, proposed problems.
This is the drama of many students, or even professors, they think you can do the stuff of Ramanujan (or similar stuff) without working extremely hard, without spending a lot of time in the reserach area.
No, you can't!
I wanna see one that contradicts me ... (I'm preparing then some integrals by Ramanujan (or similar stuff) ...)
I think behind the work of all great mathematicians like Ramanujan there is extremely hard work. His wife, as in a documentary, said Ramanujan was calculating series all day long. She didn't say 1 hour, 2 or 10, but all day long.
I work on stuff even when I go jogging, go shopping, but less when I'm tutoring or have some accounting stuff to do.
But even then I arrange in mind some things about integrals, series and limits.
Hard work can be beneficial-certainly more productive than never trying at all. And some hard endeavors are worth more than less hard ones.
But moments of insight, or simple contemplation, or just innate talent can also yield important results, and should not be discounted just because they weren't hard to achieve.
I gather you like to integrate, it fascinates and delights you.
I doubt that simple innate talent is enough for getting results like the ones of, say, Euler. I think the very hard work cannot be avoided. It's a must in my opinion.
Surely sometimes you bang your head against the wall, all the hard calculation yields nothing. Other times, the right idea just comes to you and what seemed hard is easy.
Doing hard things breeds self-discipline. Self-discipline is something you can fall back on when inspiration fails you, it's a good quality to have. It's not necessarily the only good quality worth having
Maybe not head-bashing hard work, but the labor of knowing the subject in sufficient depth (especially in seeing which problems are worth solving at all)
As I said before, hard work is "reliable"....usually doing hard work yields something of value. Some problems (hard ones, even) refuse to yield to brute force attacks.
Some problems require creativity, and I daresay creativity does not always entail "hard work".
@DavidWheeler I wonder if the good ideas always come in place without having experience in a certain field, I mean experience gained by a lot of hard work.
@MikeMiller I think I finally understood cellular boundary maps!
Let $d_2 : \Bbb Z \to \Bbb Z^{2g}$ in the cellular chain complex of $\Sigma_g$. By the boundary formula, $d_2(e^2) = \sigma_{i} d_{2i} e^1_i$.
$d_{2i}$ is the degree of the map $S^1 \to S^1$ given by attaching the boundary of $e^2$ with the $2g$-boquet, and punching everything except the $e_i^1$ petal to a point.
But since the attaching is followed according to the word $\prod [a_k, b_k]$, deleting everything else except $a_i$ leaves the word $a_i a_i^{-1}$, and attaching according to this word just gives something homotopic to the constant map.
Hence, $d_{2i}$ is always $0$, which in turn implies $d_2 = 0$.
Could someone help me better understand special limits please? For instance, limit as x goes to zero (sin(2x)/sin(3x)). Clearly I need to multiply by ((1/2x)/(1/3x)) but then what?
If you can rewrite your expression as a quotient of terms which have finite limits, then you can write the limit of the quotient as the quotient of those limits
(With the usual proviso that the denominator not go to zero)
Given a field $F$ with $q$ elements such that $q \equiv 1 \pmod n$, the equation $x^n = \alpha$ for $\alpha \in F^*$ is claimed to have $0$ or $n$ solutions. Moreover it is claimed that if $\alpha = 1$, there are no solutions. I am unable to make sense of this, surely $x = 1$ is a solution?
(I.e., if $q = 4$ and $n = 3$, then $x^3 = 1 \pmod 4$ has the solution $x = 1$
