bio | website | math.berkeley.edu/~sramesh |
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location | Berkeley, CA | |
age | 28 | |
visits | member for | 4 years, 2 months |
seen | 2 days ago | |
stats | profile views | 3,360 |
I am a graduate student in the Logic program at Berkeley, broadly interested in categorical logic and foundations of mathematics, as well as in applications of category theory to the semantics of programming languages.
Apr 9 |
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Why the underlying function of a monomorphism may not be an injection
What more are you looking for, beyond "A generalization may not always behave exactly the same as the thing it generalized, in all respects"? For what it's worth, in the quite common case of a concrete category in which the underlying set functor is representable, (i.e., in which there is a free object on one element), all monomorphisms will be injections. In some sense, the failure of monomorphisms to be injections more generally is just the failure of the underlying set functor to always be representable. |
Feb 19 |
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Nontrivial question about fibonacci numbers?
Tilings using 1x1 and 1x2 tiles? Bah! It's a direct observation that the "right parents" of each diagonal comprise the previous diagonal, while the "left parents" comprise the twice-previous diagonal. |
Feb 12 |
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Covering the Rationals — A Paradox?
So the problem is mainly your second bullet point, but it does not involve a new uncountable infinity (and what would it mean to be an uncountable infinity smaller than $\aleph_0$?). You are simply wrong in supposing that the number of gaps will be less than the number of rationals; you have no means of constructing a partial surjection from the latter to the former. |
Feb 12 |
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Covering the Rationals — A Paradox?
Indeed, instead of using mini-intervals, we might imagine removing single points: removing one rational at a time from [0, 1], we end up with n + 1 many connected components left after the first n many rationals have been removed. But after removing every rational, we are not left with a countable collection of connected components; instead, we are left with an uncountable collection of single points (the irrationals). |
Feb 12 |
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Covering the Rationals — A Paradox?
It is not true that the number of gaps must be countable. Your argument is simply "The number of gaps when the first n mini-intervals have been placed is <= n + 1; therefore, the number of gaps when all the mini-intervals have been placed is countable". But this argument is fallacious. |
Feb 9 |
awarded | Yearling |
Jan 13 |
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Direct proof of irrationality?
Slight clarification: POSITIVE integers are products of powers of primes in which the exponents are natural numbers, and POSITIVE rational numbers are products of powers of primes in which the exponents can be any integer. |
Jan 13 |
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Any example of a non-strong monad?
This interpretation will produce a non-strong monad on a CCC, if you add in all the rest of the rules of the simply-typed lambda calculus with pairs. Indeed, this produces the free monad on a CCC. |
Jan 11 |
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Any example of a non-strong monad?
Oh, whoops, nevermind A); apparently, it was corrected 5 minutes before I said it. |
Jan 11 |
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Any example of a non-strong monad?
More to the point, perhaps, depending on how you think about monads in functional programming: for a non-strong monad, one does not have a function bind : m a -> (a -> m b) -> m b in the programming language. Rather, there will be a method of taking any function of type a -> m b and turning it into a function of type m a -> m b, but this method is not carried out by any higher-order function. |
Jan 11 |
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Any example of a non-strong monad?
In functional programming, what this amounts to is that you will have to look outside the monads/functors for which one has a function map : (a -> b) -> (m a -> m b). There will in fact be a way (external to the programming language) to take any function of type a -> b and turn it into the corresponding function of type m a -> m b, but there will not be any higher-order function (internal to the programming language) which does this for you. [This is the meaning, in this context, of Finn's statement that strong monads are those which respect the internal hom's enrichment] |
Jan 11 |
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Any example of a non-strong monad?
A) How can the initial monoidal category equipped with a monad be empty? Surely, it must have an identity object for the monoidal structure. B) This is nice, as a free example with monoidal structure, but the question-asker seemed to want specifically cartesian product structure. |
Jan 11 |
asked | “Highly balanced” periodic functions |
Dec 11 |
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Are there any “related rates” calculus problems that don't feel contrived?
Hm... why is the "flow", in this particular sense, uniform? I mean, surely if I were to keep pumping a load of solid bricks into a pipe at a constant rate, ending in a vertical drop, the bricks, when dropping, would simply fall straight down, not attempting to contract inwards to maintain a uniform "flow" as they accelerate. (I may be completely missing something still) |
Dec 10 |
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Are there any “related rates” calculus problems that don't feel contrived?
I was actually looking at that same quote quite recently, and musing that I actually don't know what makes water running out of a faucet grow narrower. What is the physical principle I should be using to guide the math? |
Dec 9 |
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Proof of infinitude of primes whose reversal in base 10 is also prime
The OP is not asking about palindromes which are primes, and that is why the linked mathworld page is irrelevant. The OP is asking about primes whose reverses are also primes, regardless of whether the number and its reverse are the same. |
Nov 29 |
awarded | Good Answer |
Nov 23 |
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About a weakening of “union axiom” on ZF set theory
[Note that this is essentially just Pare's argument for the construction of coproducts from products, with the axiom of replacement slapped on at the end. The key is that ZF' and its extensions still form topoi; they are just topoi which lack the particular large cardinal consequences of the synergy of the axioms of union and replacement] |
Nov 23 |
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About a weakening of “union axiom” on ZF set theory
I believe ZF' proves that the union of two sets always exists (at least, on what I think is a natural interpretation of what ZF' is to be): Given sets $A$ and $B$, ZF' can establish the existence of $P(A) \times P(B)$ [powersets being given directly by an axiom and cartesian products requiring only the axiom of pairing (on top of extensionality)]. From this, we can use Separation to extract the subset of such pairs with one component an empty set and the other a singleton (in either A or B); finally, we can apply replacement to this to obtain the union of A and B. |
Nov 21 |
answered | Union of a object (a set) in the Elementary Theory of the Category of Sets |