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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. $\endgroup$ – Hagen von Eitzen Jul 21 '13 at 14:55
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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}? $\endgroup$ – HyperGroups Jul 21 '13 at 14:57
  • $\begingroup$ @HyperGroups Yes. $\endgroup$ – Vishal Gupta Jul 21 '13 at 15:01
  • $\begingroup$ How about an open interval $(0,1)$ and $f=0$ then $(0,1)$ is closed? $\endgroup$ – HyperGroups Jul 21 '13 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 $T_1$ $\endgroup$ – Dominic Michaelis Jul 21 '13 at 15:12
  • $\begingroup$ @HyperGroups $(0,\,1)$ is closed in the domain of $f$. $\endgroup$ – Daniel Fischer Jul 21 '13 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 '18 at 16:07

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