How to show that a set is closed?

Question

Let $f : [0, 1] \to \mathbb{R}$ be a continuous function. Let $\mathcal{S} = \{x \in [0, 1] : f(x) = 0\}$.

Show that $\mathcal{S}$ is closed.

I wonder if there is a definite way to show that a set is closed? I have tried to use the closed set definition but the answer seems too weak?

Answer 1

The two most common ways are:

1.) Show that it's complement is open (the definition)

2.) Show that every convergent sequence of points in the set has it's limit point in the set. (This way is frequently by contradiction)

Answer 2

Because f is continuous, if $C$ is closed in $\mathbb{R}$, $f^{-1}(C)$ is closed in $[0,1]$. Indeed, $\{ 0\}$ is closed in $\mathbb{R}$, so that $f^{-1}(\{0\})=S$ is closed in $[0,1]$.

Note: $\{ 0\}$ is closed in $\mathbb{R}$ because $\mathbb{R}-\{0\}=(-\infty,0)\cup (0, \infty)$, which is clearly open in $\mathbb{R}$.

Answer 3

$S$ is closed if and only if it's complement is open. $S^c=\{x\in[0,1]:f(x)\neq 0\}$. For this set to be open, that would mean that for every $x\in S^c$ there exists a ball around $x$, $B$, such that $\forall y\in B, y\in S^c$. In terms of functions, this means that near every non-root there are a lot of other non-roots. This should line up quite easily with your definition of continuous.

Answer 4

Since the existing answers use the topological definition for continuous functions, let me give an answer using the definition in elementary analysis.

Let $(x_n)_n$ be a sequence in $S$ which converges to $x \in \Bbb R$. Then for all $n$, since $x_n\in S$, $f(x_n) = 0$. Since $x_n \to x$ as $n \to \infty$, by continuity of $f$, $f(x) = 0$. Since $x_n \in S$, $0\le x_n \le 1$. Take $n \to \infty$ to see that $0 \le x \le 1$, so $x \in S$. This shows that $S$ is closed.