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This is a "think piece" about accumulation points. There are lots of _ things we might want to say about them, and it's tricky to figure out _ what we're going to need elsewhere, what's obvious, and what needs to _ be proven. So this is a collection of thoughts ... ---- So let's suppose we have a bounded sequence. We've seen that a sequence that's unbounded above has an infinite, strictly increasing sub-sequence. Since the original sequence is bounded above, so is this sub-sequence. We can ask, then, what's the *smallest* possible upper bound? What is the Least Upper Bound, or LUB. When we look at the LUB of a sequence there are two possibilities. Either the sequence actually hits that upper bound, or not. If it doesn't hit that LUB then something interesting happens: We can find a strictly increasing sub-sequence that gets closer and closer to the LUB. You say how close you want to get, and this sub-sequence gets closer than that. !* Theorem: !* Given a sequence that's bounded above and which does not achieve its LUB, there is a strictly increasing sub-sequence that gets arbitrarily close to the LUB. !* Proof: !* The first idea is this. Pick any term from the original sequence. The sequence never achieves its LUB, so that term cannot be the LUB. There must be a term that's still larger. Pick that, then lather, rinse repeat. The only problem is that the LUB of this new sequence might not be the same LUB. [[[>50 Challenge problem: create a sequence that's bounded above but for which the above process might find a subsequence with a different LUB. ]]] But we can fix that with a small twist. Here's a version that ~works. Start with any term of the original sequence. Then instead of just picking any bigger term, specifically choose a term that gets us more than half way to the LUB. We need to show that it's possible to do that. Suppose our latest term is $x,$ and suppose $b$ is the LUB. Consider $t=(x+b)/2.$ That's smaller than $b$ so it can't be an upper bound. That means there's a bigger term. Choose that as our next new term, and then repeat the process. Now we can see that each time we better than halve the distance to the LUB, so for any margin, we always end up getting closer. !/ End of Proof !/ So we've seen that if the LUB is not actually achieved then we can get a sub-sequence that gets closer and closer to the LUB. In some sense the sequence "piles up" against the LUB. If the sequence is bounded below then the same is true of the Greatest Lower Bound, again, provided the GLB is not actually achieved. The interesting thing is that given a bounded sequence there is always some point where the sequence "piles up" and gets closer and closer. There may be more than one, and it may not be the "end points" (because they may be hit once and then never approached again) but there is always at least one. "Are you sure?" - I hear you ask. !* Theorem: !* For any bounded sequence there is a point such that terms of the sequence get closer and closer. !* Proof: !* Let's play a game. You tell me how close we need to get, and I'll show you infinitely many terms of the sequence inside that limit. Suppose you choose some tiny distance $\epsilon_0.$ Remember, the sequence is bounded, so all the terms of the sequence live inside some limited space. Divide up that space into lots and lots of tiny pieces, each smaller than $\epsilon_0$ There are only finitely many pieces, so at least one of them must have infinitely many terms in it. Now change your mind and choose a smaller distance, $\epsilon_1.$ I take the already ~really small piece that has infinitely many points in it and chop it up smaller, into pieces smaller than $\epsilon_1.$ Again, there's only finitely many of them, so at least one must have infinitely many terms in it. And so we go on. You keep specifying a smaller distance, and I keep finding a place smaller than that with infinitely many terms of my sequence inside of it. So we end up with nested pieces, each one containing infinitely many terms of the original sequence. Now choose any point from the first piece. Then pick a different point from the next piece, then one from the next, and so on. [[[>50 CHALLENGE: Create a sequence where it is not true. ]]] At first glance you might think that these points get closer and closer together, but that's not always true. However, if you specify some small distance, eventually all the points will be inside a disk of that size. !/ End of Proof !/