@geocalc33 it's nice to see you working with the primes now as themselves - I.e. you're another victim of the primes and possibly the twin primes - like me :)
No matter what approach you take, it always seems to wind up on the analytical side of things, so that's why it's more of an analytic number theory problem rather than algebraic. So in other words I think you're on the right track
if f is assumed increasing and N starts at 1, the inf is f(1) and the sup is f(2). because this connects to the given information, i'm guessing that maybe f being increasing is part of the hypotheses.
ok, so the inf is f(1) = 1 and the sup is f(2) = 3.
actually i guess i'm also assuming f is continuous. without that the sup could be anything between f(1) and f(2) (inclusive), while the inf would be 1 in any event.
pick any value c between f(1) and f(2). draw a straight line from (1, f(1)) to (2,c). define f(x) by using that function as the graph for all 1 <= x < 2 and define f(2) = 3.
will be monotone increasing, and lim_{x to 2-} f(x) will be c, and not f(2) (unless you chose c = f(2))
the key is that f(1) is in that set of values of f, but f(2) doesn't have to be. it has to contain values of f(x) with x arbitrarily close to 2, but need not contain f(2).
a saving grace of plain text is that its ugliness causes people to write less of it. they focus on the essentials. a lot of old sci.math posts are still very readable.
once mathjax enters the equation, everyone can become the sorceror's apprentice, conjuring up paragraphs of perfectly typeset math babble
It's still funny that Joe Harris produced his algebraic geometry book with Machias, the word processor math fonts that Dave Bayer had championed. But it looks so horrible.
sci.math and sci.math.research. most people on those at that time were using some kind of email-like client with plaintext support only. at least i was.
Does anyone understand why a Binary Tree traversal would be more efficient for sorting large data sets than other algorithms? Like insertion, for instance?
insertion sort takes O(n) to move the next element into place (scan them all, swap min to its appropriate position) and then needs to run that once per element totalling O(n) * O(n)
if by binary tree traversal you mean heap sort, it's fairly straightforward to see why heapification is N log N, which is the best time complexity we have for sorting
i had a 'study group' of sorts but we realized early in the first semester there was really no point to it. we would get together for dinner and talk about anything other than law school.
i replaced my wheel and went north on the bay trail, ended up getting a puncture near rosie the riveters! today was not meant to be a cycling day. the ghost of willamete...
same. i hate giving them interest-free loans but i also hate having it on my to-do list.
there's probably a generation of accountants for whom the importance of timing is something of a lost art or a textbook thing. i should ask my accountant friend about that.
with interest rates doing who knows what and inflation, people are going to get good at this again.
@Osmium it has been quite a while since i dealt with such things but (assuming constant pressure) the change in enthalpy is just the heat given off or absorbed by the system?
If you can't prove that a Turing machine doesn't halt (and can't prove that it does), then there's a model of your axioms in which it does halt. This model looks like the set $\Bbb N$ union many "extra" numbers that are larger than everything in $\Bbb N$. In a sense, this means that, though the Turing machine doesn't halt in the "standard model" $\Bbb N$, you just haven't waited long enough!
Nonstandard models of arithmetic are weird.
In fact, there's a "universal" Turing machine such that, for any partial function $\subseteq\Bbb N\to\Bbb N$, there exists a model of arithmetic in which it will compute exactly that function
And if the domain of the partial function isn't all of $\Bbb N$, for any partial function extending the domain of that function to more inputs, there's an extension of the model in which it computes the extended function
i have a question about why vakil's "magic diagram" is cartesian - actually my concern is found in this comment (i am not convinced by the response): math.stackexchange.com/questions/778186/…
i think the issue i'm having is that if we have a morphism $Y\rightarrow Z$ and we draw out the big diagram involving $X_1\times_{Y}X_2\rightarrow X_1\times_{Z}X_2$, i don't know how to show this diagram actually commutes (and maybe it doesn't, but it seems like we need this for the argument given in the above post to work)
more specifically i don't see any reason why the compositions $X_1\times_{Z}X_2\rightarrow X_i\rightarrow Y$ should be equal, if they even necessarily are - all we know is that they are equal after composing with $Y\rightarrow Z$
Its in relation to a description of a polyhedral which I believe it makes sense to talk about equalities being elementwise, but I dont have a firm grasp of the difference between them.
@P-addict they certainly aren't in general, but the argument doesn't need that either
like, think you are in Sets and take $Z$ to be a singleton. then the map between the fiber products becomes the canonical inclusion $X_1\times_YX_2\subseteq X_1\times X_2$. of course, the projections from the product to the factors followed by the maps to $Y$ are not equal, unless the inclusion is an equality
in fact, generally, if they were equal, you'd obtain a map between the fiber products the other way round and this would be inverse to the map already given, since the composites are just canonical projections in each component
@Thorgott right, this too... i knew something was up when i was getting results like "all fibered products of $X_1,X_2$ are isomorphic"
any idea on how to show the diagram is cartesian? i think the yoneda lemma is one option but as was mentioned in the post this exercise shows up in vakil's notes before he mentions yoneda so i was looking for another proof
@Thorgott oh! is it that the blue and red paths are equal because once we get to $Y\times_{Z}Y$ we can compose with the map $Y\times_{Z}Y\rightarrow Y$, and since the last map in both the blue and red paths to $Y\times_{Z}Y$ is $Y\rightarrow Y\times_{Z}Y$, we get the composition $Y\rightarrow Y\times_{Z}Y\rightarrow Y=\text{id}_Y$?
I'm trying to prove that $|\mathbb Z| = |\mathbb Z - \{2\}|$. I know one way to do this is to show that there is a bijection between the sets. Can I do this by forming the function $f: \mathbb Z - \{2\} \rightarrow \mathbb Z$, such that $f(z) = \frac{1}{2-z}$ and then showing the bijection?
Another option I could think of: Since $\mathbb Z$ is denumerable, it forms a bijection with the natural numbers. So I can just claim this, and then show that $\mathbb Z - \{2\}$ also has a bijection with the natural numbers and therefore $|\mathbb Z| = |\mathbb Z - \{2\}|$.
under: your rule for f doesn't produce integer outputs for most z, but you could fix it so that it does. the second approach doesn't seem much simpler than the first (although sometimes with other sets and contexts it might be)
under: e.g. it should be possible to prove very quickly that f(z) = z if z < 2 and z - 1 if z > 2 defines a bijection from Z - {2} to Z
@P-addict this is true, but I don't see how this is relevant. the blue and the red path are both the composition $X_1\times_YX_2\rightarrow X_1\rightarrow Y$.
by the second approach, i generally mean, showing that A and B have the same cardinality via some chain of bijections A -> C_1 -> C_2 -> ... -> C_n -> B which are known to exist, or are more simply described or proved to be bijections, than a 'directly' constructed map from A to B (e.g. because an explicit formula would be annoying to compute)
@Thorgott well, i guess what i'm saying is when trying to prove the universal property, we start off with two maps through $Y$ and $X_1\times_{Z}X_2$ from an object $A$ which are equal when they go to $Y\times_{Z}Y$. it turns out both compositions actually come from mpas which first go through the $X_i$ then $Y$ (namely the blue and red paths) and end with $Y\rightarrow Y\times_{Z}Y$, so since this map is mono we actually have maps into $Y$ which play nicely with the $X_i\rightarrow Y$...
...and that is how we get our map $A\rightarrow X_1\times_{Y}X_2$, by the universal property
in other words i was stuck on showing the red and blue paths were equal, and this follows from the fact that they are once we continue them to $Y\times_{Z}Y$ via a mono map
I found an even simpler solution: $\mathbb Z - \{2\}$ is an infinite subset of a denumerable set, $\mathbb Z$. There is a theorem that states that any such subset is also denumerable, and as such, $|\mathbb Z - \{2\}| = |\mathbb Z|$.
the morphism $A\rightarrow X_1\times_ZX_2$ already determines the two morphisms $A\rightarrow X_1$ and $A\rightarrow X_2$ that will determine the morphism $A\rightarrow X_1\times_YX_2$, the additional data merely specifies the compatibility necessary for this morphism to be induced
I don't quite follow you, but the fact that that map is a mono is not particularly relevant here
i don't agree that this is simpler. you are detouring through two facts (that Z - 2 is infinite, and that Z is denumerable) which are useful generalities but have their own proofs somewhere in terms of bijecitons, when you could just be exhibiting one function and proving it is a bijection.
a lot of general facts about cardinalities are proved via cardinal arithmetic, which is basically what you are doing here. but cardinal arithmetic can be subtle with infinite sets, and it's arguably overkill when you can just write a map down.
@Thorgott i think the thing i wasn't sure on was why the two morphisms $A\rightarrow X_i$ we got agreed when composed with the $X_i\rightarrow Y$. they agree when you then compose this with $Y\rightarrow Z$ (obviously), but that they do so before was what i needed to show
all statements about cardinal arithmetic (e.g. |A| = |B| because |A| = |C| and |C| = |B|) are encoding statements about bijections that you aren't bothering to write down. this is fine when that's the shortest route and the maps you would be writing down have complicated formulas or involve tons of arbitrary choices.
i mean, it's generally fine, but as a matter of aesthetics or simplicity, explicit bijections beat cardinal arithmetic
you see this a whole lot even in finite sets in combinatorics. often it is far more informative to have a bijection between two sets of things than simply to know those sets have the same size
yeah, you have to use that the maps to $Y\times_ZY$ agree and then look at them component-wise (because that's how we understand maps into fiber products usually)
that's also implicit in your observation that $Y\rightarrow Y\times_ZY$ is a mono
e.g. there are the same number of k-element subsets of X = {1,2,...,n} as there are (n-k)-element subsets of {1,2,...,n}. the simplest proof is to show that S -> X \ S defines a bijection between those collections of sets. you don't need to compute what that number is to do that, as other routes to establishing this fact might do
How does one complete the character table of a finite group knowing only data of conjugacy classes? Only way I could think of is via induced representations
i'd almost say that in some contexts it would be missing the point to bring binomial coefficients (i.e. counting, i.e. bijections with {1,2,...,size of set}) into it
