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

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

And should the function on a closed interval?

• Hint: The inverse image of a closed set is closed. Jul 21, 2013 at 14:55

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

• Should this mean that {0,0,0,0,,,infinity} is simply {0}? Jul 21, 2013 at 14:57
• @HyperGroups Yes. Jul 21, 2013 at 15:01
• How about an open interval $(0,1)$ and $f=0$ then $(0,1)$ is closed? Jul 21, 2013 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 $T_1$ Jul 21, 2013 at 15:12
• @HyperGroups $(0,\,1)$ is closed in the domain of $f$. Jul 21, 2013 at 15:31

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 $f$.

• How do you know that there must be a converging sequence $\{x_n\}$ in the zero set? Oct 1, 2018 at 16:07