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Interchanged $f$ with $x$
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edit approved Apr 9, 2023 at 16:32
Nitin Uniyal
edit approved Apr 9, 2023 at 16:32
Nitin Uniyal
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There isn't such a function. If $f$ is nonzero and continuous at some point $x$, then there is a neighbourhood of $f$$x$ on which $x$$f$ doesn't vanish. Hence if the zero set of zeros of $f$ is dense, then the function has to be discontinuous at every value on which it doesn't vanish.

There isn't such a function. If $f$ is nonzero and continuous at some point $x$, then there is a neighbourhood of $f$ on which $x$ doesn't vanish. Hence if the zero set of $f$ is dense, then the function has to be discontinuous at every value on which it doesn't vanish.

There isn't such a function. If $f$ is nonzero and continuous at some point $x$, then there is a neighbourhood of $x$ on which $f$ doesn't vanish. Hence if the set of zeros of $f$ is dense, then the function has to be discontinuous at every value on which it doesn't vanish.

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Wojowu
Wojowu
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There isn't such a function. If $f$ is nonzero and continuous at some point $x$, then there is a neighbourhood of $f$ on which $x$ doesn't vanish. Hence if the zero set of $f$ is dense, then the function has to be discontinuous at every value on which it doesn't vanish.