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I wish to find out the class of compact sets in $\mathbb{R}$ which are supports of continuous functions.

To my solution I have an idea that the support will be finite union of closed and bounded sets in $\mathbb{R}$ but I am not sure about the answer and if it is correct then i have no explicit form to prove it. can anyone help me in doing so.

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    If "support" means "image is non-zero", then your functions apparently are real-valued, and as $\mathbb{R}\setminus\{0\}$ is an open set, so is $f^{-1}(\mathbb{R}\setminus\{0\})$ for all continuous functions $f$. So the only compact support of a continuous function would be the empty set. – 2017-02-15
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    Support is per definition the closure of "image is non-zero". – 2017-02-15
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    I think your argument is not true as for example take $f(x)=\begin{cases}(e^x-1)*(e^x-e) & \text{if}\,\,\,x\in[0,1] \\ 0 & \text{if}\,\,\, otherwise\end{cases}$. then here function is continuous and it's support is [0,1] – 2017-02-15
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    By definition, the support of a continuous function is the closure of an open set. Then it follows that $S = \operatorname{cl}(\operatorname{int}(S))$. And if $U$ is an open subset of a metric space $(X,d)$, then $$f \colon x \mapsto \operatorname{dist}(x, X\setminus U) = \inf \{ d(x,y) : y \notin U\}$$ is a continuous function whose support is $\operatorname{cl}(U)$. Thus in a metric space, a closed set is the support of a continuous function if and only if it is the closure of an open set; if and only if it is the closure of its interior. – 2017-04-26

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