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Prove: the set of zeros of a continuous function is closed.

And should the function on a closed interval?

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    $\begingroup$ Hint: The inverse image of a closed set is closed. $\endgroup$ Jul 21, 2013 at 14:55

2 Answers 2

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Hint: No matter what the domain of the function is (as long as it has a topology — no restriction on "closed interval", in particular):

  • The inverse image of an open set by a continuous function is open.
  • The inverse image of a closed set by a continuous function is closed.

and you're looking for $f^{-1}(\{0\})$.

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    $\begingroup$ Should this mean that {0,0,0,0,,,infinity} is simply {0}? $\endgroup$ Jul 21, 2013 at 14:57
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    $\begingroup$ @HyperGroups Yes. $\endgroup$ Jul 21, 2013 at 15:01
  • $\begingroup$ How about an open interval $(0,1)$ and $f=0$ then $(0,1)$ is closed? $\endgroup$ Jul 21, 2013 at 15:11
  • $\begingroup$ For functions going from a topological space to a $T_1$ space this is always true, but please notice that $\{0\}$ doesn't need to be close if your space isn't $T_1$ $\endgroup$ Jul 21, 2013 at 15:12
  • $\begingroup$ @HyperGroups $(0,\,1)$ is closed in the domain of $f$. $\endgroup$ Jul 21, 2013 at 15:31
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This depends on how you define continuity. If you say that a function $f$ is continuous if and only if $f^{-1}(U)$ is open for any open set $U$, then you can show that this is equivalent to $f^{-1}(V)$ being closed for any closed set $V$, and then we're done by taking $V=\{0\}.$

If instead you define continuity in the sense of preserving limits, take a sequence $\{x_n\}$ contained in the zero set (i.e. so that $f(x_n)=0$), which converges to some number $x.$ Then $f(x)=f(\lim x_n)=\lim f(x_n)=\lim 0=0,$ so $x$ is in the zero set, showing that the zero set contains all of its limit points and is then closed. Here it's in the statement $f(\lim x_n)=\lim f(x_n)$ that we're using continuity of $f$.

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    $\begingroup$ How do you know that there must be a converging sequence $\{x_n\}$ in the zero set? $\endgroup$
    – Aqqqq
    Oct 1, 2018 at 16:07
  • $\begingroup$ @Aqqqq, we're proving that if there is a sequence of points of the zero set which is convergent, then its limit is automatically an element of the zero set, as well. However, unless the zero set is empty, there will always be a sequence of points of the zero set which is convergent. Do you see why? $\endgroup$ Aug 21, 2023 at 22:30

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