Orthonormal Hamel Basis is equivalent to finite dimension

Question

Consider a Hilbert space which is infinite dimensional. If it is separable, it is well known that an orthonormal basis will be countable, while a hamel basis will be uncountable (since it is a complete space), therefore the two bases cannot coincide. But how can we prove that those two bases cannot be equal if our space is not separable?

Answer 1

The argument can be made rather simpler: if you have a given ON basis $e_i$, $i\in I$, the vector $v=\sum_{n\in {\bf N}} 2^{-n}e_{i_n}$, where $i_n$ are distinct, is clearly not a finite linear combination of the $e_i$ (because if we subtract from $v$ some finite linear combination of $e_i$, we can still find some $i_n$ such that the result is not orthogonal to $e_{i_n}$).

You don't need $I$ to be countable.

On the other hand, if you have a Hilbert space of Hilbert dimension at least $\mathfrak c$, then the Hilbert dimension and Hamel dimension do coincide (but with different bases). They also coincide with the cardinality of the space, in this case.