Confusion in the proof of essential spectrum of toeplitz operator $T_\phi$ is connected for $\phi\in H^\infty+C(\Bbb{T})$ I am reading this theorem from Banach Algebra Techniques in Operator Theory by Douglas (Corollary 7.37). The proof goes as follows-

Here essential spectrum of $T_\phi$ is the set $\sigma_e(T_\phi)=\{\lambda\in \Bbb{C}:\ T_{\phi-\lambda}\text{ is not fredholm}\}$.
It is proved that $T_{\phi-\lambda}$ is fredholm if and only if $\phi-\lambda$ is invertible in $H^\infty+C(\Bbb{T})$. I also know that $\psi\in H^\infty+C(\Bbb{T})$ is invertible in $H^\infty+C(\Bbb{T})$ if and only if $\exists \epsilon>0,\ \delta\in(0,1)$ such that $|\hat{\psi}(re^{it})|\ge\epsilon$ for all $r\in(1-\delta,1),\ e^{it}\in\Bbb{T}$ here $\hat{\psi}$
denotes the harmonic extension of $\psi$ defined as $\hat{\psi}(re^{it})=\sum\limits_{n\in\Bbb{Z}} \hat{\psi}(n)r^{|n|}e^{int}$.
Using these two results $\lambda\in\sigma_e(T_\phi)\iff\forall\epsilon>0,\ \delta\in(0,1)$ there exists $r\in (1-\delta,1),\ e^{it}\in\Bbb{T}$ such that $|\hat{\phi}(e^{it})-\lambda|<\epsilon\iff\forall\delta\in(0,1),\ \lambda\in\overline{\{\hat{\phi}(re^{it}):\ r\in(1-\delta,1),\ e^{it}\in\Bbb{T}\}}\iff\lambda\in \bigcap\limits_{\delta\in(0,1)}\overline{\{\hat{\phi}(re^{it}):\ r\in(1-\delta,1),\ e^{it}\in\Bbb{T}\}}$
Now since $\hat{\phi}$ is continuous, image of the annulus $\{re^{it}:\ r\in(1-\delta,1),e^{it}\in \Bbb{T}\}$ will be connected and hence the closure. Therefore, each of the above set is connected. But how to show that the intersection is connected, in general intersection is not connected. The book says, "Since each of these sets is connected, the result follows". I cannot understand how it follows.
Can anyone help me in this regard? Thanks for your help in advance.
 A: What you have is a monotone family of connected compact sets, so the intersection is connected.
Suppose that $\{E_\delta\}_{\delta\in(0,1)}$ is a family of compact connected sets and $E_\delta\subset E_{\delta'}$ if $\delta'>\delta$. Let $E=\bigcap_\delta E_\delta$. The set $E$ is nonempty, being a decreasing union of compact.
Suppose that  $E=F\cup G$, with $F,G$ closed and disjoint. Because $F,G$ are compact, there exist disjoint open sets $V,W$ with $F\subset V$ and $G\subset W$.
Claim: there exists $\delta$ such that $E_{\delta}\subset V\cup W$.
Assuming the claim, $E_\delta=(E_\delta\cap V)\cup(E_\delta\cap W)$. As $E_\delta$ is connected, either $E_\delta\subset V$ or $E_\delta\subset W$. The former case implies $E\subset V$, which in turn implies $E=F$. The other case similarly implies $E=G$. Thus $E$ is connected.
Proof of the Claim. Suppose that $E_\delta\subset V\cup W$ fails for all $\delta$. This would mean that $E_\delta\cap (V\cup W)^c\ne\varnothing$ for all $\delta$. As our family is monotone and $(V\cup W)^c$ is closed, this implies that the family $\{E_\delta\cap(V\cup W)^c\}$ is a family of compact sets with the finite intersection property. This then implies that
$$
\varnothing\ne \bigcap_\delta E_\delta\cap(V\cup W)^c=E\cap(V\cup W)^c.
$$
But this contradicts $E\subset V\cup W$. In conclusion there has to exists $\delta$ such that $E_\delta\cap(V\cup W)^c=\varnothing$; that is, $E_\delta\subset V\cup W$.
