$ \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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Image of Compact Subset is Compact

Suppose $K$ is a compact subset of $\mathbb{R}^n$ and $f:K\to\mathbb{R}^m$ is a continuous function on $K$. Then $f(K)$ is compact.

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Image of Compact Subset is Compact

Suppose $K$ is a compact subset of $\mathbb{R}^n$ and $f:K\to\mathbb{R}^m$ is a continuous function on $K$. Then $f(K)$ is compact.

Concepts

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Hypothesis

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Proof

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FullPage
result
concepts
hypothesis
implications
proof

Image of Compact Subset is Compact

Suppose $K$ is a compact subset of $\mathbb{R}^n$ and $f:K\to\mathbb{R}^m$ is a continuous function on $K$. Then $f(K)$ is compact.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon

Image of Compact Subset is Compact

Suppose $K$ is a compact subset of $\mathbb{R}^n$ and $f:K\to\mathbb{R}^m$ is a continuous function on $K$. Then $f(K)$ is compact.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon
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concepts
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proof