TheoremBase

Suprema and Finite Sums in [0,∞]

In [0,∞], suprema commute with adding or multiplying by a constant and with sums of nondecreasing sequences; finite sums are monotone, homogeneous, and commute with suprema of nondecreasing families.

Statement

In the setting of Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices, let [0,∞][0,\infty] and the order ≤\le be as in The Extended Real Line, Its Order, and [0,∞] (ii) and The Extended Real Line, Its Order, and [0,∞] (iii), the sum ++ and the product ⋅\cdot on [0,∞][0,\infty] as in Addition and Multiplication on [0,∞] (i) and Addition and Multiplication on [0,∞] (ii), and suprema and sup⁡n\sup_{n} as in Order and Suprema in the Extended Real Line (iii). Let c∈[0,∞]c\in[0,\infty] and let XX be a nonempty subset of [0,∞][0,\infty].

sup⁡{x+c:x∈X}=sup⁡X+c\sup\{x+c:x\in X\}=\sup X+c.

sup⁡{cx:x∈X}=csup⁡X\sup\{cx:x\in X\}=c\sup X.

If (un)(u_{n}) and (vn)(v_{n}) are sequences in [0,∞][0,\infty] with un≤un+1u_{n}\le u_{n+1} and vn≤vn+1v_{n}\le v_{n+1} for every n∈Nn\in\mathbb{N}, then sup⁡n(un+vn)=sup⁡nun+sup⁡nvn\sup_{n}(u_{n}+v_{n})=\sup_{n}u_{n}+\sup_{n}v_{n}.

Let AA be a finite set. Sums over AA of maps into [0,∞][0,\infty] are as in Sums and Products over a Finite Set and over an Interval (i), Sums and Products over a Finite Set and over an Interval (ii), Sums and Products over a Finite Set and over an Interval (iii) and Sums and Products over a Finite Set and over an Interval (v), for the operation ++, which is associative and commutative with neutral element 00 by Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i).

If f,g:A→[0,∞]f,g:A\to[0,\infty] satisfy f(y)≤g(y)f(y)\le g(y) for every y∈Ay\in A, then ∑y∈Af(y)≤∑y∈Ag(y)\sum_{y\in A}f(y)\le\sum_{y\in A}g(y).

If f:A→[0,∞]f:A\to[0,\infty] and B⊆AB\subseteq A, then ∑y∈Bf(y)≤∑y∈Af(y)\sum_{y\in B}f(y)\le\sum_{y\in A}f(y).

If f:A→[0,∞]f:A\to[0,\infty], then c∑y∈Af(y)=∑y∈Acf(y)c\sum_{y\in A}f(y)=\sum_{y\in A}cf(y).

∑y∈A0=0\sum_{y\in A}0=0.

If w:N×A→[0,∞]w:\mathbb{N}\times A\to[0,\infty] satisfies w(n,y)≤w(n+1,y)w(n,y)\le w(n+1,y) for all n∈Nn\in\mathbb{N} and y∈Ay\in A, then ∑y∈Asup⁡nw(n,y)=sup⁡n∑y∈Aw(n,y)\sum_{y\in A}\sup_{n}w(n,y)=\sup_{n}\sum_{y\in A}w(n,y).

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