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Let $f \colon \mathbb{R}^2 \to \mathbb{R}$ be a continuous function. Let $D$ be the closed unit disc in $\mathbb{R}^2$. Is $f(D)$ necessarily an interval in $\mathbb{R}$? If it is an interval, which of the forms $(a, b)$, $[a, b)$, $(a, b]$ and $[a, b]$, with $a, b\in\mathbb{R}$ can it have?

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Do you know about connectedness and compactness? –  Nils Matthes Sep 10 '12 at 13:07
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$f[D]$ is connected and compact (as these are properties preserved by continuous images), and the only such sets in $\mathbb{R}$ are sets of the form $[a,b]$ where $a \le b$.

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$D$ is connected and $f$ is continuous. Then $f(D)$ is connected. The connected sets in $\mathbb{R}$ are the intervals and as $D$ is compact $f(D)$ is bounded and closed. Hence $f(D) = [a,b]$.

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