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

And should the function on a closed interval?

Sigur
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HyperGroups
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2 Answers2

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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 $f^{-1}(\{0\})$.

Clement C.
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    Should this mean that {0,0,0,0,,,infinity} is simply {0}? – HyperGroups Jul 21 '13 at 14:57
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    @HyperGroups Yes. – Vishal Gupta Jul 21 '13 at 15:01
  • How about an open interval $(0,1)$ and $f=0$ then $(0,1)$ is closed? – HyperGroups Jul 21 '13 at 15:11
  • 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 $T_1$ – Dominic Michaelis Jul 21 '13 at 15:12
  • @HyperGroups $(0,,1)$ is closed in the domain of $f$. – Daniel Fischer Jul 21 '13 at 15:31
  • Yep — for the induced topology on $I=(0,1)$, on which $f\colon I \to \mathbb R$ is defined. The topology $\mathcal T$ is defined in terms of open sets $O=I\cap O^\prime$, for $O^\prime$ an open set of $\mathbb R$; and the closed sets are all sets of the form $I\setminus O$, for $O\in\mathcal T$. So $I$ itself is (as it should be) both an open and a closed set. – Clement C. Jul 21 '13 at 16:43
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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 $f$.

youler
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    How do you know that there must be a converging sequence ${x_n}$ in the zero set? – Aqqqq Oct 01 '18 at 16:07
  • @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? – Cameron Buie Aug 21 '23 at 22:30