Hilbert space, Quantum mechanics, Completeness. As we know, Hilbert space $\mathbb{H}$  is defined as the complete inner product space. In here, the completeness means the convergence of each Cauchy sequence in
$\mathbb{H}$. 
However, some QM texts states the completeness of Hilbert space  as follows:
The space $\mathbb{H}$ is complete if and only if
given orthonormal subset(countable) spans $\mathbb{H}$.
Now, I have a question about them.
Is there any relation between two complete view points? 
Are they equivalent? 
 A: Short answer: no. The first definition is the definition of completeness. The second definition is a characterisation of a different property: separability. The more common way of defining a separable space is one with a countable, dense subset. Here's an incomplete space that's countable:
$$c_{00} = \lbrace (x_1, x_2, x_3, \ldots x_n, 0 , 0, \ldots) \in l^2(\mathbb{N}) : n \in \mathbb{N} \rbrace$$
You can see that the standard basis $e_i$, which contains only $0$s except for a $1$ in the $i$th place, form an orthonormal basis for the space. Alternatively, you can form a countable dense subset by taking all sequences of the form $(q_1, q_2, \ldots, q_n, 0, 0, \ldots)$ where $q_1, \ldots, q_n \in \mathbb{Q}$. It's not complete, because you can form a Cauchy sequence by
\begin{align*}
\left(1, 0, 0, 0, 0, 0, \ldots\right) \\
\left(1, \frac{1}{2}, 0, 0, 0, 0, \ldots\right) \\
\left(1, \frac{1}{2}, \frac{1}{3}, 0, 0, 0, \ldots\right) \\
\left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, 0, 0, \ldots\right)
\end{align*}
etc, that doesn't converge in $c_{00}$ (because it converges to an element of $l^2(\mathbb{N}) \setminus c_{00}$.
On the other hand, take $l^2(\mathbb{R})$, which is the set of functions from $\mathbb{R}$ to $\mathbb{R}$ (or $\mathbb{C}$) which has countable support and the sum of the (moduli of the) square of the function values converges. The inner product is defined by, for $f, g \in l^2(\mathbb{R})$,
$$\langle f, g \rangle = \sum_{x \in \mathbb{R}} f(x)g(x) = \sum_{x \in \mathrm{supp}(f) \cap \mathrm{supp}(g)} f(x)g(x).$$
Such an inner product space is complete, but not separable. To see it's not separable, for $r \in \mathbb{R}$, let $e_r \in l^2(\mathbb{R})$ be the characteristic function for $\lbrace r \rbrace$, i.e. map $r$ to 1 and all else to $0$. Then, the family of open balls $B(e_r, 1/\sqrt{2})$ indexed by $r \in \mathbb{R}$ is uncountable and pairwise disjoint. Any dense subset must intersect with each of these balls, and must do so at different points in each of the uncountable balls, meaning that the dense set must not be countable.
Completeness isn't quite so neat, but you can see it easily enough by embedding the sequence back into the known complete space $l^2(\mathbb{N})$. If you have a sequence of functions from $\mathbb{R}$ with countable supports, then the unions of the countable supports is countable itself, meaning that we can map the non-zero terms of all the sequences isometrically back onto $l^2(\mathbb{N})$, where it will still be Cauchy, and still have a limit, which we can recover back to $l^2(\mathbb{R})$.
Long story short, neither condition necessarily implies the other.
A: The first statement is the correct definition. The second statement is a theorem for separable Hilbert spaces.
