Before you get banned, can you tell me why $\lim_{k \to \infty} \bigg[ \big(f(x_1)-f(x_1-1/k)\big)\bigg(\frac{x-(x_1-1/k)}{x_1-(x_1-1/k)}\bigg)+f(x_1-1/k)\bigg]=f(x)$? :S
So we're doing linear approximation by secants approaching the tangent line at $x_1$, @Twink. If $f$ is differentiable at $x_1$, those approach the tangent line. But there's no reason $f(x)$ should have anything to do with what the tangent line at $x_1$ is.
You definitely have a workable idea. You want to "bridge the gap" where $f$ has a discontinuity, and your approach is good EXCEPT for using $x_1$ as one of your points.
You know that's the correct formula for a linear function joining the two points. So you're ok. When you posted your original question, you didn't specify where you wanted your $x$ to be. That's why I dismissed it. You only want $x$ to be near and to the left of $x_1$.
the argument we just talked through ON THE LEFT works also on the RIGHT :)
Think about the definition of limit, not the formula.
OK, I need to get out of here. Plus, my blood is still boiling at being flagged and deleted for giving intelligent advice but making it clear I was annoyed at the person for ignoring me. I've learned my lesson.
@Twink: Here's an analogous problem. Think about the sequence of functions that start at $(0,0)$, go up to $(1/2k,1)$, back down to $(1/k,0)$ and then are $0$ all the way to $x=1$. What is the pointwise limit? Proof?
@KarlKronenfeld it was something our TA gave us to help prove or disprove the limit rules. This is just something that defines convergence in upper limit topology
@NikolajKyed The rule is indeed still valid in the upper limit topology. You have to say that you can always choose a point between $b_n$ and $a_n$ (in this order) for sufficiently large $n$.
@KarlKronenfeld Then I will assume I can apply the 3rd lemma because both a_n and b_n converge in the ordinary topology and because I can always choose a point between b_n and a_n for sufficiently large n then it will still be valid in the upper limit topology. Is this correct?
@KarlKronenfeld So it's enough to say I can choose a point between b_n and a_n for sufficiently large n for which the rule will still be valid? Which makes the lemmas useless in this case?
@KarlKronenfeld So because I can choose this point it makes the rule valid? As for the first lemma it's only true if a_n and b_n converge in the upper limit topology, but the thing is I only know they converge which does not necessarily mean they converge in the upper limit topology.
@KarlKronenfeld Okay they both converge in the upper limit and therefore also in the ordinary. For 5 you said I could always choos a point between b_n and a_n for sufficiently large n. Following that with the statement from the first lemma the rule is therefore valid in this topology?
@NikolajKyed ok, things are easier to see when you write down the relevant continuous map. Maybe revisit this problem when you learn about them and see if you can find the map I am referring to.
@KarlKronenfeld Okay thanks for the help, I'm not sure if I can construct a proper argument for 5 but I assume my TA will tell me what I did wrong if I have to redo the assignment. For number 3 you said something about it not always being true for the termwise product. In class we concluded that in the lower limit topology it was not correct, but he never explained in detail why. Any hints?
@KarlKronenfeld that did trigger something in my brain. I'm not sure what you mean by an increasing sequence of negative real numbers. Do you mean a sequence {-3,-2,-1}?
@KarlKronenfeld Oh yea now I think I remember. because 0 is not contained in the sequence it won't converge. At least he made an example in lower limit topology wher -1/n does not converge and 1/n does converge
@KarlKronenfeld Yes I noticed that one question was to prove or disprove that 1/n converges to 0 in upper limit topology and my answer was exactly like yours. But wouldn't that be the same as to say that the product still converges if both are greater than 0?
Lemma 2: "If a sequence has a decreasing subsequence (moves to the left), then the sequence does not converge in the upper limit topology"
Lemma 3: "If a sequence converges in the original topology and overall it is increasing (moves to the right), then it converges in the upper limit topology"
Okay that made it very clear. the product is the subsequence. So if the subsequence is increasing it will indeed converge in upper limit topology. Correct?
So because the original sequences move to the right and the product move to the left lemma 2 states they won't converge in upper limit topology? Sorry if I seem confused but I really want to understand what's happening.
Forgive me if I'm wrong: For lemma 3 |x_n > x| < infty and because a_n and b_n converge in the usual topology for which a_n and b_n < a and b respectively then a_n and b_n converges in upper limit topology. However if a and b < 0 then lemma 2 states that a_n * b_n does not converge in the ordinary topology.
Okay thanks, sorry for the confusion but I want to understand this. I should be able to figure out 5 now after what we just discussed. Have a good evening or day wherever you are now from :)
Anyone familiar with Bernoulli Space? Ie. a sequence of either 1s or 0s (coin tossing). I am wondering does this space only include infinite sequenes of 1s and 0s such as 1, 1, 0, 0, 1, ...? Or does it also include finite sequences of 1s and 0s...such as the set sequence 1, 0, 0, 0, 1?
