# Separable Hilbert space have a countable orthonormal basis

I want to show that every an infinite-dimensional separable (contains countable dense set) Hilbert space has a countable orthonormal basis.

I know that every orthogonal set in a separable Hilbert space is countable,it is help me with the proof?

This is actually an if and only if statement. For future students looking for a full answer to this question, I am posting a full proof below:

Let $$H$$ be an infinite-dimensional Hilbert space. Show that $$H$$ has a countable orthonormal basis if and only if $$H$$ has a countable dense subset.

First let us assume that $$H$$ has a countable orthonormal basis $$\{e_i\}$$. Then any $$x$$ can be uniquely written as $$x=\sum\limits_{i=1} c_ie_i \quad \text{where} \quad c_i= \langle x,e_i \rangle$$ Recall that $$S:=\mathbb{Q}+\mathbb{Q}i$$ is a countable dense subset of $$\mathbb{C}$$. Now for every $$n \in \mathbb{N}$$, consider the following subset of $$H$$: $$A_n=\sum\limits_{i=1}^n s_ie_i \quad \text{where} \quad s_i \in S \, \, \forall i$$ Being a finite union of countable sets, each $$A_n$$ is countable. Then define $$A:= \bigcup_{n=1} A_n$$ Being a countable union of countable sets, $$A$$ is countable. Let us show that $$A$$ is the desired dense subset. Let $$x\in H$$. Let $$\epsilon >0$$ be arbitrary. Then $$x=\sum\limits_{i=1} c_ie_i \quad \text{for some} \quad c_i \in \mathbb{C}$$ Since this sum is convergent in the norm, we can find $$N$$ big enough such that $$\left\| \sum\limits_{i=N+1} c_ie_i\right\|< \frac{\epsilon}{2}$$ Also, Since $$S$$ is a dense subset of $$\mathbb{C}$$, for every $$i \leq N$$, we can find $$s_i$$ such that $$|c_i-s_i|<\frac{\epsilon}{2^{i+1}}$$ Now consider the following element $$x_N=\sum\limits_{i=1}^N s_ie_i \in A$$ We know that $$\sum\limits_{i=N+1} c_ie_i$$ is the difference of two elements of $$H$$ and thus is in $$H$$ and therefore by using triangle inequality and Perseval's inequality, we have $$$$\begin{split} \|x-x_N\| & =\left\|\sum\limits_{i=1} c_ie_i-\sum\limits_{i=1}^N s_ie_i\right\|\\ & =\left\|\sum\limits_{i=1}^N (c_i-s_i)e_i+\sum\limits_{i=N+1} c_ie_i\right\|\\ & \leq \left\|\sum\limits_{i=1}^N (c_i-s_i)e_i\right\|+\left\| \sum\limits_{i=N+1} c_ie_i\right\|\\ & \leq \sum\limits_{i=1}^N \left|(c_i-s_i)\right|+\frac{\epsilon}{2}\\ & < \sum\limits_{i=1}^N \frac{\epsilon}{2^{i+1}}+\frac{\epsilon}{2}\\ & \leq \sum\limits_{i=1}\frac{\epsilon}{2^{i+1}}+\frac{\epsilon}{2}\\ & = \epsilon \\ \end{split}$$$$ Therefore, $$x_n$$ is an element of the set $$A$$ that is in the ball of radius $$\epsilon$$ around $$x$$. Since both $$x$$ and $$\epsilon$$ were arbitrary, this implies that $$A$$ is dense in $$H$$.

Conversely, assume that $$H$$ has a countable dense subset $$\{a_j\}$$, where $$j \in \mathbb{N}$$. Let $$\{e_i\}_{i \in I}$$ be an orthonormal basis of $$H$$ (we know that such a basis exists by Zorn's lemma). For contradiction, assume that the orthonormal basis is uncountable. Then notice that for any $$e_n \neq e_m$$, $$n,m \in I$$, by perpendicularity we have $$$$\begin{split} \|e_n-e_m\|^2 & =\langle e_n-e_m, e_n-e_m \rangle\\ & =\langle e_n, e_n \rangle+\langle e_m, e_m \rangle-2\text{Re}\big(\langle e_n, e_m \rangle\big)\\ & =\|e_n\|+\|e_m\|\\ & =2\\ \end{split}$$$$

Therefore, any two elements of our orthonormal basis are $$\sqrt{2}$$ apart. Now for all $$i \in I$$, consider the following balls: $$B\left(e_i, \frac{1}{2}\right)$$ Each of such balls has diameter less than $$1$$. Thus, each one of them can contain at most one element of the basis, namely only the center itself. Also, these ball are disjoint, since if there exists an element in two balls, then by the triangle inequality the distance between the centers of the balls is less than $$1$$, which is a contradiction. Since $$\{a_j\}$$ is a dense subset, it has to have at least one element in each such ball. Since by the above remarks the balls are disjoint, we have a surjective function from some $$\{a_j\}$$ to the balls. However, our balls are indexed by an uncountable set. Thus, there has to be at least an uncountable amount of $$a_j$$, a contradiction. Thus, any orthonormal basis has to be countable.

• Line 4 is wrong. If your terms were squared, then it'd be Parseval's, but that follows from triangle inequality instead. I'm referring to your statement $$\left\| \sum_{i=1}^N (c_i - s_i) e_i \right\| = \sum_{i=1}^N \left\| (c_i - s_i) \right\|$$ which should instead be $$\left\| \sum_{i=1}^N (c_i - s_i) e_i \right\| \leq \sum_{i=1}^N \left\| (c_i - s_i) \right\|.$$ – Glassjawed Jun 18 '17 at 0:33
• You're correct. I changed the equality sign to the inequality sign. Thank you. – Pawel Jun 20 '17 at 7:23

Hint: Take a countable dense subset $Q$ and build an orthonormal basis of $\text{span}(Q)$.

• Build by Gram–Schmidt process ? – ali baba Mar 8 '13 at 9:20
• How am I define Q ?: – ali baba Mar 8 '13 at 9:26
• @ali baba: $Q$ is any countable dense subset of $\mathcal{H}$, which exists because you have assumed that $\mathcal{H}$ is separable. – Haskell Curry Mar 8 '13 at 9:32

You can use Zorn's lemma to show that there exists a maximal orthogonal set, show that it has to be a Schauder basis. Then use what you know about orthogonal sets.

(Remark: Zorn's full power is not needed here. It's an overkill, but it's easy and convenient.)

The above hint with more details: Every Hilbert space has an orthonomal basis - using Zorn's Lemma

• If I assume that Hilbert space contains a uncountable ortonormal basis and shows that I get a conflict, is that acceptable? – ali baba Mar 9 '13 at 21:56
• It isn't enough. Bases are not only orthonormal, but they are maximal. It might be the case that every orthonormal set is countable, but there is a point which is not in its closure. You need to show that there is a set which is not only orthonormal, but its span is dense. – Asaf Karagila Mar 9 '13 at 21:59