Continuous functions can be represented by their rational support; in particular, for each real number r, choose a sequence rn of rational numbers converging to r, and let f(r):=limn→∞f(rn).
Therefore, there is an injection from the vector space of continuous functions C to the vector space of all sequences c: since the rationals are countable, enumerate them by (r1,r2,…). Then the sequence (f(r1),f(r2),…) represents continuous function f.
This map is not a surjection because not every map from the rational numbers to the real numbers is continuous, and so not every sequence represents a continuous function. It is injective, and so it shows that a basis for the latter space is at least as large in cardinality as a basis for the former space. One can construct an injective map in the other direction, showing the both spaces of bases with the same cardinality, and so they are isomorphic.
Continuous functions can be represented by their rational support; in particular, for each real number r, choose a sequence rn of rational numbers converging to r, and let f(r):=limn→∞f(rn).
Therefore, there is an injection from the vector space of continuous functions C to the vector space of all sequences c: since the rationals are countable, enumerate them by (r1,r2,…). Then the sequence (f(r1),f(r2),…) represents continuous function f.
This map is not a surjection because not every map from the rational numbers to the real numbers is continuous, and so not every sequence represents a continuous function. It is injective, and so it shows that a basis for the latter space is at least as large in cardinality as a basis for the former space. One can construct an injective map in the other direction, showing the both spaces of bases with the same cardinality, and so they are isomorphic.
Fixed, thanks.