What compact means
A bounded operator K on a Banach space is compact if it maps the unit ball to a set with compact closure — equivalently, every bounded sequence (xₙ) has a subsequence for which (Kxₙ) converges. Finite-rank operators are compact, and a norm-limit of compact operators is compact, so compactness is the closure of finite-rank operators in operator norm. Compactness is precisely the property that recovers Bolzano–Weierstrass behaviour in infinite dimensions, where the unit ball is otherwise never compact.
The Riesz–Schauder spectral picture
Here is the payoff. For a compact operator K on an infinite-dimensional space, the spectrum σ(K) is a countable set with 0 as its only possible accumulation point. Every nonzero point of σ(K) is an eigenvalue of finite multiplicity, and for each ε > 0 only finitely many eigenvalues satisfy |λ| ≥ ε. The spectrum is therefore a sequence of eigenvalues λₙ → 0 (possibly finite), plus the point 0 itself, which always belongs to σ(K) in infinite dimensions because K cannot be invertible.
Why eigenvalues cannot accumulate away from 0.
Suppose lambda_n -> lambda with |lambda| > 0, distinct eigenvalues,
with eigenvectors e_n. Distinct eigenvalues => the e_n are independent.
Let M_n = span(e_1, ..., e_n), a strictly increasing chain of closed subspaces.
By Riesz's lemma pick y_n in M_n with ||y_n|| = 1 and dist(y_n, M_{n-1}) >= 1/2.
For m < n, compute K y_n - K y_m:
(K - lambda_n) maps M_n into M_{n-1}, and K y_m lies in M_{n-1}, so
K y_n - K y_m = lambda_n y_n - [ stuff in M_{n-1} ].
Hence ||K y_n - K y_m|| >= |lambda_n| * dist(y_n, M_{n-1}) >= |lambda_n|/2.
Since lambda_n -> lambda != 0, ||K y_n - K y_m|| >= |lambda|/4 for large n,m.
So (K y_n) has NO convergent subsequence, contradicting compactness of K.
Therefore eigenvalues can only accumulate at 0. QEDThe Fredholm alternative
For a compact K and λ ≠ 0, the operator K − λI behaves like a matrix in the cleanest sense: it is a Fredholm operator of index zero. Concretely, the Fredholm alternative says exactly one of two things holds. Either K − λI is invertible (so the equation (K − λI)x = y has a unique solution for every y), or λ is an eigenvalue, in which case the homogeneous equation (K − λI)x = 0 has a nonzero solution and the inhomogeneous equation is solvable only for y orthogonal to the finite-dimensional kernel of the adjoint. There is no middle ground — no continuous spectrum away from 0.