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."

-

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Trevor Wilson, Daniel Rust, Nick Peterson, Stefan Hamcke, M Turgeon
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $\{x \in [a,b]: h(x) = 0 \} = h^{-1}(\{0\})$.

-

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.

-
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 $x\in T$.

-