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Math Help - Dimension of all complex valued continuos functions

  1. #1
    Member Ruun's Avatar
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    Dimension of all complex valued continuos functions

    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 n \to \infty. (Thinking as the function as a vector written as a linear combination of the elements of the basis)

    Thanks in advance.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Take into account that the family:

    \mathcal{F}=\left\{{f_n:\mathbb{R}\rightarrow{\mat  hbb{C}},\;f_n(x)=x^n,\;n\in \mathbb{N}}\right\}

    is linearly independent.

    Regards.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Ruun View Post
    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 n \to \infty. (Thinking as the function as a vector written as a linear combination of the elements of the basis)

    Thanks in advance.
    In fact, if \dim_{\mathbb{C}}C\left[\mathbb{R},\mathbb{C}\right]=2^{\aleph_0}. To see this we merely note that we may inject C\left[\mathbb{R},\mathbb{C}\right] into \mathbb{C}^{\mathbb{Q}} via the map f\mapsto f_{\mid \mathbb{Q}}. 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 \text{card }\mathbb{C}^{\mathbb{Q}}=\left(2^{\aleph_0}\right)  ^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0} so that \text{card }C\left[\mathbb{R},\mathbb{C}\right]\leqslant 2^{\aleph_0} and in particular if \mathcal{F} is a basis for C\left[\mathbb{R},\mathbb{C}\right] then \text{card }\mathcal{F}\leqslant 2^{\aleph_0}. But, note that the family of maps \left\{f_r: r\in\mathbb{R}^+\right\} given by f_r:\mathbb{R}\to\mathbb{R}:x\mapsto x^r is linearly independent and has cardinality 2^{\aleph_0} and so \text{card }\mathcal{F}\geqslant 2^{\aleph_0}. It follows by the Schroeder-Bernstein theorem that \text{card }\mathcal{F}=\dim_{\mathbb{C}}C\left[\mathbb{R},\mathbb{C}\right]=2^{\aleph_0}
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  4. #4
    Member Ruun's Avatar
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    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) \text{card }\mathcal{F}\leqslant 2^{\aleph_0} \mbox{ and} \text{ card }\mathcal{F}\geqslant 2^{\aleph_0} \mbox{ then } \text{card }\mathcal{F}=2^{\aleph_0} right? And as we know the cardinal of the basis we know the dimension.

    Now the notation: I don't know if \mathbb{C}^{\mathbb{Q}} stands for the cartesian product of \mathbb{C} , \mathbb{Q} times ("the number of rationals times").

    Also I don't know wich set is f_{\mid \mathbb{Q}}.

    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.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Ruun View Post
    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).
    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

    First, let's see if I understood your post:

    I think that the idea is that the theorem states (for our particular case) \text{card }\mathcal{F}\leqslant 2^{\aleph_0} \mbox{ and} \text{ card }\mathcal{F}\geqslant 2^{\aleph_0} \mbox{ then } \text{card }\mathcal{F}=2^{\aleph_0} right? And as we know the cardinal of the basis we know the dimension.
    Correct. I am appealing to something called the Schroeder-Bernstein Theorem which in essence says that the relation \leqslant on cardinal numbers (where X\leqslant Y means "There exists an injection f:X\to Y" 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 2^{\aleph_0} then \dim_{\mathbb{C}}C[\mathbb{R},\mathbb{C}]=2^{\aleph_0}

    Now the notation: I don't know if \mathbb{C}^{\mathbb{Q}} stands for the cartesian product of \mathbb{C} , \mathbb{Q} times ("the number of rationals times").
    In essence, yes. But, with the definition of Cartesian product extended to the arbitrary case; namely that if \left\{X_{\alpha}\right\}_{\alpha\in\mathcal{A}} is an indexed class of sets then

    \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=\left\{f:\mat  hcal{A}\to\bigcup_{\alpha\in\mathcal{A}}X_{\alpha}  :f(\alpha)\in X_{\alpha}\text{ for all }\alpha\in\mathcal{A}\right\}

    So, without all the jargon really \mathbb{C}^{\mathbb{Q}} is the set of all functions from \mathbb{Q} into \mathbb{C}.


    Also I don't know wich set is f_{\mid \mathbb{Q}}.
    This is just the restriction of f to \mathbb{Q}. Namely, for a function f:\mathbb{R}\to\mathbb{C} we can create a new function, denoted f\mid_{\mathbb{Q}} which maps \mathbb{Q}\to\mathbb{C} and such that f_{\mid\mathbb{Q}}(x)=f(x) for all x\in\mathbb{Q}. 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 \mathbb{R} which agree on \mathbb{Q} must be the same function (ask if you don't see why).
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