The finite case is fine, as a vector space it is easy to show that R^X is isomorphic to R^n.(adsbygoogle = window.adsbygoogle || []).push({});

What about when X is infinite? I believe it is true in general that dim(R^X) = #(X), which I hope holds in the infinite case too. I know that the set given by B={b_x; x in X} defined as b_x(y) = delta_xy gives a linearly independent set in R^X. But as far as I am aware, the dimension of a vector space is the cardinality of a basis. Since this is not a basis (I found by googling), it does not give the dimension of the space correct?

If not, how can one construct a basis of R^X? Though I don't care too much about what the basis is, just the size of it.

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# Questions about a basis and others for the vector space R^X (functions from X->R)

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