Notation: Let $a,b,m,M\in\mathbb{R}$, such that $a<m\leq M<b$, and let $A$ be a self-adjoint bounded operator on a Hilbert space $\mathcal{H}$ satisfying $mI\leq A\leq MI$. Let $K[a,b]$ denote the set of all piecewise continuous bounded functions which are monotone decreasing limits of continuous functions. Consider the function $$ e_\lambda(t):= \begin{cases} 1, & t\leq\lambda \\ 0, & t>\lambda. \tag{1} \end{cases} $$ Obviously $e_\lambda(t)\in K[a,b]$ and so we can define $e_\lambda(A)$. Let $E_\lambda:=e_\lambda(A)$.
My first question is the following: The way that $e_\lambda(A)$ is defined is to be the limit of polynomials $P_n(A)$, where $P_n(t)$ are polynomials that approximate (via Weierstrass approximation theorem) the sequence of functions $\varphi_n(t)\in C[a,b]$, which they themselves approximate the function $e_\lambda(t)$. That is, $\lim P_n(A):=e_\lambda(A):=E_\lambda$. But is it true that one can write $$ E_\lambda:=e_\lambda(A) = \begin{cases} I, & A\leq\lambda I \\ 0, & A>\lambda I, \tag{2} \end{cases} $$ where $I$ and $0$ are the identity and zero operators, respectively?
Next the author$^{[1]}$ gives the following example:
Example: Let $A$ be a compact self-adjoint operator: $Ax=\sum\lambda_k\langle x,e_k \rangle e_k$. Then $E_\lambda x=\sum_{\lambda_k\leq\lambda}\langle x,e_k \rangle e_k$ for $\lambda<0$ and $$ E_\lambda x= x-\sum_{\lambda_k>\lambda}\langle x,e_k \rangle e_k\tag{3} $$ for $\lambda>0$.
And to be straightforward: I don't understand this at all. It'd be helpful if someone could really break down what is going on here. For instance, it seems that $(2)$ can't be true, since then in the example above $E_\lambda x$ would only ever equal $I x$ or $0 x$, correct?
$^{[1]}$ The text is Functional Analysis: An Introduction by Eidelman-milman-tsolomitis; more specifically, chapter 7 of the text.