I am not sure -- a manifold is locally connected and has countable basis?
There is an Exercise in a book as following :
Given a Manifold M , if N is a sub-manifold , an V is open set then V \cap N is a countable collection of connected open sets .
I am asking why he put this exercise...
That I am talking about . My only question what 's the meaning of there exist x such that ... .
In the definition above it and and some of axioms of set theory for example Axiom of Extensionality we begin by saying for all x and y , ... . So what " for all " Exactly mean.
I see you mean Axiom Schema of Comprehension right . So in the definition I have written at first such domain is not determined , I see that he wants to determine any set , and I can't take such domain to be the st of all sets because it doesn't exist . And I don't want to use notion of classes.
The idea begins When Babylonians have developed a system for measuring angles "degree measure", They discovered some trigonometric theorems but not explicitly , and then the sin function was first developed in India . But trigonometry wasn't translated to Europe through the Hindus , but through...
That I am talking about , the book haven't considered classes yet ( I know some about them ) , but the question is when I say for all x such that P(x) , will it mean "if you have proved the existence of set x then P(x) holds" or what , and whwn I say there exists some x such that P(x) holds ...
But we are dealing wit general sets in set theory so we consider "any sets" also I don't know what must be meant for saying "any sets". When I say there exist x such that ... , I understand that as "x exists according to axioms of ZFC " , But When I was reading in Logic , and when defining...
What is meant by that ??
This can considered as a logic question , when I say for example there exist x such that ... . Mustn't I define some set from which x belong .
In a book of set theory it defined a binary relation as following :
A set R is a binary relation if (\forall x \in...
ok that is was I tried to show but the step I was stuck in that if we differentiate x^n , ntimes we will have n! , so I don't know if I have proved it in the right way , I had done this by induction , where if n=1 , we will have (x^1)' = 1 = 1! , and hence we will assume that this true for any k...
Suppose fn:[0,p/q] → ℝ be a function defined by :
f_{n}(x) = \frac{x^{n}(1-qx)^{n}}{n!} where p,n and q are natural numbers .
Is that true that f_{n}^{(2n)} is always an integer for any natural number n .
Thanks .
No , but in his analysis book he says that in principle changing co-domains changes the function . and I see that it is not necessarily the case , that is for example if B ≠ f(A) m doesn't mean that these two functions are not equal f : A → B and f : A → f(A) .