Theorem: Spectrum of Self-Adjoint Densely-Defined Linear Operator is Real and Closed

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a self-adjoint densely-defined linear operator.
Let $\map \sigma T$ be the spectrum of $\struct {\map D T, T}$. 
Then $\map \sigma T$ is a closed subset of $\C$ and: 
:$\map \sigma T \subseteq \R$

Proof:
Let $\lambda \in \map \sigma T$.
We show that $\lambda \in \R$. 
From Element of Spectrum of Densely-Defined Linear Operator not in Residual Spectrum is Approximate Eigenvalue, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ with:
:$\paren {T - \lambda I} x_n \to 0$
with $\norm {x_n} = 1$ for each $n \in \N$. 
Then, we have: 
{{begin-eqn}}
{{eqn	| l = \cmod {\innerprod {\paren {T - \lambda I} x_n} {x_n} }
		| o = \le
		| r = \norm {\paren {T - \lambda I} {x_n} } \norm {x_n}
		| c = Cauchy-Bunyakovsky-Schwarz Inequality
}}
{{eqn	| r = \norm {\paren {T - \lambda I} {x_n} }
		| c = since $\norm {x_n} = 1$
}}
{{eqn	| o = \to
		| r = 0
		| c = Modulus of Limit: Normed Vector Space
}}
{{end-eqn}}
So:
:$\innerprod {\paren {T - \lambda I} x_n} {x_n} \to 0$
We have: 
{{begin-eqn}}
{{eqn	| l = \innerprod {\paren {T - \lambda I} {x_n} } {x_n}
		| r = \innerprod {T x_n - \lambda x_n} {x_n}
}}
{{eqn	| r = \innerprod {T x_n} {x_n} - \lambda \innerprod {x_n} {x_n}
}}
{{eqn	| r = \innerprod {T x_n} {x_n} - \lambda \norm {x_n}^2
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| r = \innerprod {T x_n} {x_n} - \lambda
}}
{{end-eqn}}
So, from Sum Rule for Complex Sequences, we have: 
:$\innerprod {T x_n} {x_n} \to \lambda$ 
From Convergence of Complex Conjugate of Convergent Complex Sequence, we have: 
:$\overline {\innerprod {T x_n} {x_n} } \to \overline \lambda$
We then have: 
{{begin-eqn}}
{{eqn	| l = \overline {\innerprod {T x_n} {x_n} }
		| r = \innerprod {x_n} {T x_n}
		| c = conjugate symmetry of the inner product
}}
{{eqn	| r = \innerprod {T x_n} {x_n}
		| c = {{Defof|Self-Adjoint Densely-Defined Linear Operator}}
}}
{{end-eqn}}
So:
:$\innerprod {T x_n} {x_n} \to \overline \lambda$
From Convergent Complex Sequence has Unique Limit, we have: 
:$\lambda = \overline \lambda$
From Complex Number equals Conjugate iff Wholly Real, we then have: 
:$\lambda \in \R$
{{qed|lemma}}
We now show that $\map \sigma T$ is closed.
We show that $\C \setminus \map \sigma T$ is open.
That is, we show that the resolvent set of $T$, $\map \rho T$, is open.
Let $\lambda \in \map \rho T$.
Then $T - \lambda I$ is injective, has everywhere dense image and $\paren {T - \lambda I}^{-1}$ is bounded.
So there exists a real number $C > 0$ such that: 
:$\norm {\paren {T - \lambda I}^{-1} y} \le C \norm y$ for each $y \in \map {\paren {T - \lambda I} } \HH$.
Each $y \in \map {\paren {T - \lambda I} } {\map D T}$ can be written $y = {\paren {T - \lambda I} } x$ for $x \in \map D T$ and conversely $y$ of this form is contained in $\map {\paren {T - \lambda I} } {\map D T}$.
Then we have: 
:$\norm {\paren {T - \lambda I}^{-1} \paren {T - \lambda I} x} \le C \norm {\paren {T - \lambda I} x}$ for each $x \in \map D T$
That is: 
:$\ds \norm {\paren {T - \lambda I} x} \ge \frac 1 C \norm x$ for $x \in \map D T$.
We show that $\map \rho T$ contains an open neighborhood of $\lambda$.
Let $\mu \in \C$ have: 
:$\ds \cmod {\lambda - \mu} < \frac 1 {2 C}$
Then, for each $x \in \map D T$ we have: 
{{begin-eqn}}
{{eqn	| l = \norm {\paren {T - \mu I} x}
		| r = \norm {\paren {T - \lambda I} x - \paren {\lambda - \mu} x}
}}
{{eqn	| o = \ge
		| r = \norm {\paren {T - \lambda I} x} - \norm {\paren {\lambda - \mu} x}
		| c = Reverse Triangle Inequality
}}
{{eqn	| o = \ge
		| r = \frac 1 C \norm x - \cmod {\lambda - \mu} \norm x
}}
{{eqn	| o = \ge
		| r = \frac 1 C \norm x - \frac 1 {2 C} \norm x
}}
{{eqn	| r = \frac 1 {2 C} \norm x
}}
{{end-eqn}}
Then, if $\paren {T - \mu I} x = 0$ for some $x \in \map D T$, we have $\norm x = 0$, so $x = 0$. 
So $T - \mu I$ is injective.
Suppose that $\mu \in \map \sigma T$. 
From Partition of Spectrum of Densely-Defined Linear Operator, we have that $\mu$ is contained in either the continuous spectrum of $T$ or the residual spectrum of $T$. 
From Self-Adjoint Densely-Defined Linear Operator has Empty Residual Spectrum, the residual spectrum of $T$ is empty, so if $\map \sigma T$, we must have that $\mu$ is in the continuous spectrum of $T$.
Then $\paren {T - \mu I}^{-1}$ is not bounded, but we have, for each $y \in \map {\paren {T - \mu I} } {\map D T} = \map D {\paren {T - \mu I}^{-1} }$ we have: 
{{begin-eqn}}
{{eqn	| l = \norm y
		| r = \norm {\paren {T - \mu I} \paren {T - \mu}^{-1} y}
		| c = setting $x = \paren {T - \mu I}^{-1} y$ in the previous computation
}}
{{eqn	| o = \ge
		| r = \frac 1 {2 C} \norm {\paren {T - \mu I}^{-1} y}
}}
{{end-eqn}}
so:
:$\norm {\paren {T - \mu I}^{-1} y} \le 2 C \norm y$
for all $y \in \map D {\paren {T - \mu I}^{-1} }$.
So, we have that $\paren {T - \mu I}^{-1}$ is bounded, a contradiction.
So we have $\mu \not \in \map \sigma T$, so $\mu \in \map \rho T$. 
So for all $\mu$ with: 
:$\ds \cmod {\lambda - \mu} < \frac 1 {2 C}$
we have $\mu \in \map \rho T$.
So $\map \rho T$ is open.
So $\map \sigma T$ is closed.
{{qed}}