Take into account that the family:
is linearly independent.
Regards.
Hi!
I'm trying to show that the set of all complex valued continuous functions defined on the real line is an infinite-dimensional complex vector space.
Well I don't know where to start.. My first intuitive idea is to show that if we take the Taylor series of the function only up to n finite terms this sum will not be equal to the function, only in the limit . (Thinking as the function as a vector written as a linear combination of the elements of the basis)
Thanks in advance.
In fact, if . To see this we merely note that we may inject into via the map . This is an injection since if two continuous maps into a metric space (in fact, a Hausdorff space) agree on a dense subset they must be the same function. But, note that so that and in particular if is a basis for then . But, note that the family of maps given by is linearly independent and has cardinality and so . It follows by the Schroeder-Bernstein theorem that
Thanks for even showing the dimension of the space.
I'm having some trouble with your notation, I'm a completely noob on real analysis. (My motivation of studying this is to study the Hilbert Spaces of Quantum Mechanics, wich are not quite good covered at school and I think that we are missing a huge point here).
First, let's see if I understood your post:
I think that the idea is that the theorem states (for our particular case) right? And as we know the cardinal of the basis we know the dimension.
Now the notation: I don't know if stands for the cartesian product of , times ("the number of rationals times").
Also I don't know wich set is .
I guess this is just above my math prerequisites, and I understand that an answer that I can understand I not just a post on a forum. (several weeks even months of study).
Thanks for your time.
Absolutely no worries! Notation can sometimes make math unreadable, and I in particular have a penchant for using "fancy looking notation" just to look "fancy". But anways
Correct. I am appealing to something called the Schroeder-Bernstein Theorem which in essence says that the relation on cardinal numbers (where means "There exists an injection " is antisymmetric). And then, right again. I'm appealing to the Dimension Theorem (which, by the way, is not trivial to prove. It requires something called the ultrafilter lemma) to conclude that since the cardinality of any basis must have cardinality thenFirst, let's see if I understood your post:
I think that the idea is that the theorem states (for our particular case) right? And as we know the cardinal of the basis we know the dimension.
In essence, yes. But, with the definition of Cartesian product extended to the arbitrary case; namely that if is an indexed class of sets thenNow the notation: I don't know if stands for the cartesian product of , times ("the number of rationals times").
So, without all the jargon really is the set of all functions from into .
This is just the restriction of to . Namely, for a function we can create a new function, denoted which maps and such that for all . In other words, we just make the domain "smaller" and consider the function on that "smaller set". It turns out, like I said, that two continuous functions on which agree on must be the same function (ask if you don't see why).Also I don't know wich set is .