= Solution
For a bounded operator on a complex <Banach space>, the <resolvent set> is $\rho(T)=\{\lambda:\lambda I-T\text{ is bijective with bounded inverse}\}$, and the <resolvent operator> is $R_T(\lambda)=(\lambda I-T)^{-1}$. The <spectrum> is its complement $\sigma(T)=\mathbb C\setminus\rho(T)$; the <point spectrum> consists of $\lambda$ with $\ker(\lambda I-T)\ne0$. For a real Banach space these definitions use its complexification.
If $|\lambda|>\|T\|$, the norm-convergent <Neumann series> gives $R_T(\lambda)=\lambda^{-1}\sum_{n\geq0}(T/\lambda)^n$, so the spectrum is bounded. If $\lambda_0\in\rho(T)$, factor
$$
\lambda I-T=(\lambda_0I-T)[I+(\lambda-\lambda_0)R_T(\lambda_0)].
$$
The second factor is invertible by a Neumann series whenever $|\lambda-\lambda_0|\|R_T(\lambda_0)\|<1$. Thus the resolvent set is open and \b[the spectrum is closed and bounded].
The point spectrum need not be closed. On $\ell^2$, define $Te_n=e_n/n$. Every $1/n$ is an <eigenvalue>, but zero is not: $Tx=0$ forces every coordinate of $x$ to vanish. No other <eigenvalues> occur, so $\sigma_p(T)=\{1/n:n\geq1\}$ is not closed.
Finally, if $T$ is self-adjoint and $Tv=\lambda v$, $v\ne0$, then $\langle Tv,v\rangle$ is real and equals $\lambda\|v\|^2$ under the convention linear in the first argument. Therefore \b[every <eigenvalue> of a self-adjoint operator is real].
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