$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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proof

The Extreme Value Theorem

Let $K$ be a nonempty compact subset of $\mathbb{R}^n$ and let $f:K\to\mathbb{R}$ be a continuous function. Then there exists $\vec{a}, \vec{b}\in K$ such that $f(\vec{a})\leq f(\vec{x})\leq f(\vec{b})$ for all $\vec{x}\in K$.

Concepts

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Hypothesis

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Results

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Proof

$K$ is compact and $f$ is continuous implies that $f(K)$ is compact which means $f(K)$ is closed and bounded.
$f(K)\subseteq\mathbb{R}$ is nonempty and bounded. Then there exists some $M=\text{sup}f(K)$ exists. By the definition of a supremum, a sequence $(y_k)_{k=1}^{\infty}$ in $f(k)$ such thta $M-\frac{1}{k} A similar argument for the infinimum.

The Extreme Value Theorem

Let $K$ be a nonempty compact subset of $\mathbb{R}^n$ and let $f:K\to\mathbb{R}$ be a continuous function. Then there exists $\vec{a}, \vec{b}\in K$ such that $f(\vec{a})\leq f(\vec{x})\leq f(\vec{b})$ for all $\vec{x}\in K$.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

$K$ is compact and $f$ is continuous implies that $f(K)$ is compact which means $f(K)$ is closed and bounded.
$f(K)\subseteq\mathbb{R}$ is nonempty and bounded. Then there exists some $M=\text{sup}f(K)$ exists. By the definition of a supremum, a sequence $(y_k)_{k=1}^{\infty}$ in $f(k)$ such thta $M-\frac{1}{k} A similar argument for the infinimum.
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result
concepts
hypothesis
implications
proof
FullPage
result
concepts
hypothesis
implications
proof

The Extreme Value Theorem

Let $K$ be a nonempty compact subset of $\mathbb{R}^n$ and let $f:K\to\mathbb{R}$ be a continuous function. Then there exists $\vec{a}, \vec{b}\in K$ such that $f(\vec{a})\leq f(\vec{x})\leq f(\vec{b})$ for all $\vec{x}\in K$.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

$K$ is compact and $f$ is continuous implies that $f(K)$ is compact which means $f(K)$ is closed and bounded.
$f(K)\subseteq\mathbb{R}$ is nonempty and bounded. Then there exists some $M=\text{sup}f(K)$ exists. By the definition of a supremum, a sequence $(y_k)_{k=1}^{\infty}$ in $f(k)$ such thta $M-\frac{1}{k} A similar argument for the infinimum.

The Extreme Value Theorem

Let $K$ be a nonempty compact subset of $\mathbb{R}^n$ and let $f:K\to\mathbb{R}$ be a continuous function. Then there exists $\vec{a}, \vec{b}\in K$ such that $f(\vec{a})\leq f(\vec{x})\leq f(\vec{b})$ for all $\vec{x}\in K$.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

$K$ is compact and $f$ is continuous implies that $f(K)$ is compact which means $f(K)$ is closed and bounded.
$f(K)\subseteq\mathbb{R}$ is nonempty and bounded. Then there exists some $M=\text{sup}f(K)$ exists. By the definition of a supremum, a sequence $(y_k)_{k=1}^{\infty}$ in $f(k)$ such thta $M-\frac{1}{k} A similar argument for the infinimum.
FullPage
result
concepts
hypothesis
implications
proof