Suppose $A subset mathbb{R}$ is closed. How can we show that there exists a continuous real-valued function $f$ such that $ker(f)=A$, i.e.e there is a function vanishing exactly on $A$?
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Suppose $A subset mathbb{R}$ is closed. How can we show that there exists a continuous real-valued function $f$ such that $ker(f)=A$, i.e.e there is a function vanishing exactly on $A$?