# Suppose $f:[a,b] \to \mathbb{R}$ and $g:[a,b] \to \mathbb{R}$ are both continuous. Let $T=\{x:f(x) = g(x)\}$. Prove that $T$ is closed. [closed]

As the title says.

Suppose $f:[a,b] \to \mathbb{R}$ and $g:[a,b] \to \mathbb{R}$ are both continuous. Let $T=\{x:f(x) = g(x)\}$. Prove that $T$ is closed.

This should be based on the definition, "A set is closed iff every accumulation point of the set belongs to the set."

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## closed as off-topic by Trevor Wilson, Daniel Rust, Nick Peterson, Stefan Hamcke, M TurgeonOct 9 '13 at 0:44

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If $h:[a,b]\rightarrow \mathbb R$ is continuous then $h$ preserves closed and open sets under taking preimages. Proving that $h$ preserves open sets is quite easy I guess. If not let me know. Closed subsets: If $V \subseteq \mathbb R$ is closed then $V^c$ is open, so $h^{-1}(V^c) = h^{-1}(V)^c$ is open. So $h^{-1}(V)$ is closed. Then look at $\{x \in [a,b]: h(x) = 0 \} = h^{-1}(\{0\})$.

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Prove that if $h$ is a continuous function, then $S = \{x : h(x) = 0\}$ is closed (Hint: What is the value of $h$ at an accumulation point?).

Now use the fact that $h=f-g$ is continuous.

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what is h referring to here? I didn't use one and I'm confused. –  Nick Sep 24 '13 at 4:46
@Nick Think about this $h$ as a function from auxilary statement before the proof of your statement –  Evgeny Sep 24 '13 at 5:06

Let $x$ be an accumulation point of $T$. Then there exists a sequence $\{x_{n}\}\subset T$ that converges to $x$. Furthermore $f(x_{n})=g(x_{n})$ for all $n$, since $x_{n}\in T$. Next $\lim f(x_{n})=\lim g(x_{n})$. Since $f$ and $g$ are continuous you may pass the limit through and we have $f(x)=g(x)$ so $x\in T$.

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