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  "body": "\n## Statement\n\nLet $\\mathbb{K}$ be $\\R$ or $\\C$, and let $A$ be a Banach algebra over $\\mathbb{K}$ with unit $1$, so that $\\norm{xy} \\le \\norm{x} \\norm{y}$ for all $x, y \\in A$.\nLet $a \\in A$.\nThen the [[spectrum|EH0CZ19RKP]] $\\sigma(a)$ is a [[compact subset|0R6EWFKRCF]] of $\\mathbb{K}$, and $\\abs{\\lambda} \\le \\norm{a}$ for every $\\lambda \\in \\sigma(a)$.\n\n## Remarks\n\nIf $\\abs{\\lambda} \u003e \\norm{a}$, then $a - \\lambda 1 = -\\lambda (1 - a/\\lambda)$, and the Neumann series $\\sum_{n \\ge 0} (a/\\lambda)^n$ converges to an inverse of $1 - a/\\lambda$.\nSo $\\sigma(a)$ lies in the closed disc of radius $\\norm{a}$.\nIf $u$ is invertible and $\\norm{b} \u003c 1/\\norm{u^{-1}}$, then $u - b = u(1 - u^{-1} b)$ is invertible by the same series.\nHence the invertible elements form an open set, the [[resolvent set|X0243CZD9D]] is open, and $\\sigma(a)$ is closed and bounded.\n\nCompleteness is necessary.\nIn the [[normed algebra|4DJ4V5VBSH]] of complex polynomials with the sup norm on $[0,1]$, only nonzero constants are invertible.\nSo the element $t$ has spectrum $\\C$, which is not bounded.\n\nFor a complex Banach algebra the spectrum is also nonempty.\nFor instance, for $a \\in M_n(\\C)$ with the operator norm, the theorem says that every eigenvalue has modulus at most $\\norm{a}$.\n\n## Explanations\n\n### L1 human\n\nThe numbers $\\lambda$ for which $a - \\lambda$ is not invertible form a closed and bounded set, and they are never larger than the size of $a$.\nThe reason is that $1 - x$ is invertible as soon as $x$ is small, by the geometric series $1 + x + x^2 + \\cdots$.\n\n### L2 human\n\nThis is the first basic fact of spectral theory in Banach algebras.\nTogether with the holomorphy of the resolvent and [[Liouville's theorem|0EHKVRF4J6]], it gives the [[nonemptiness of the spectrum|QWQC5SV0SD]] in the complex case.\nIt is refined by [[Gelfand's spectral radius formula|T4RFYT8HDN]], $\\max \\set{\\abs{\\lambda} : \\lambda \\in \\sigma(a)} = \\lim_n \\norm{a^n}^{1/n}$, which can be smaller than $\\norm{a}$.\nThe compactness allows the continuous functional calculus, which needs $\\sigma(a)$ to be a compact space.\n",
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