# Thread: prove that T is closed.

1. ## prove that T is closed.

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

2. Originally Posted by tn11631
Suppose f: [a,b] -> R and g: [a,b]-> R are both continuous. Let T={x: f(x)=g(x)}. Prove that T is closed.

Suppose $\displaystyle x_0\in [0,1]\setminus T\Longrightarrow f(x_0)-g(x_0)=c\neq 0 \Longrightarrow$ since [/tex]f(x)=g(x)[/tex] is continuous on $\displaystyle [0,1]$ we get that there exists $\displaystyle \epsilon > 0$ s.t.

$\displaystyle f(x)-g(x)\neq 0\,\,\,\forall\,x\in I_\epsilon:=(x_0-\epsilon,x_0+\epsilon)\Longrightarrow I_\epsilon\subset [0,1]\setminus T$ $\displaystyle \Longrightarrow [0,1]\setminus T$ is open ...

Tonio

3. Originally Posted by tn11631
Suppose f: [a,b] -> R and g: [a,b]-> R are both continuous. Let T={x: f(x)=g(x)}. Prove that T is closed.
Here is a second way.
Suppose that $\displaystyle \left( {t_n } \right)$ is a sequence of points in $\displaystyle T$.
If $\displaystyle \left( {t_n } \right) \to z$ then showing that $\displaystyle z\in T$ we have done the problem.

4. Originally Posted by tonio
Suppose $\displaystyle x_0\in [0,1]\setminus T\Longrightarrow f(x_0)-g(x_0)=c\neq 0 \Longrightarrow$ since [/tex]f(x)=g(x)[/tex] is continuous on $\displaystyle [0,1]$ we get that there exists $\displaystyle \epsilon > 0$ s.t.

$\displaystyle f(x)-g(x)\neq 0\,\,\,\forall\,x\in I_\epsilon:=(x_0-\epsilon,x_0+\epsilon)\Longrightarrow I_\epsilon\subset [0,1]\setminus T$ $\displaystyle \Longrightarrow [0,1]\setminus T$ is open ...

Tonio
Sorry I just want to clarify symbols. Where you have \ were using the "a-b" right? looking at it in this box i see that it says setminus which is what i thought. and then do we have to show that f(x)=g(x) is continuous on [0,1]?

5. I think the second way is a bit easier to do...

Since both f and g are continuous, you know that for a sequence $\displaystyle \{ x_n \}$ such that $\displaystyle x_n \rightarrow x_0$, we have $\displaystyle f(x_n) \rightarrow f(x_0) , g(x_n) \rightarrow g(x_0)$.

Now, think about this, if the sequence $\displaystyle x_n$ is from T, then we should have $\displaystyle f(x_n) = g(x_n)$, right? Well, you just need to prove that $\displaystyle x_0$ is also in T. In other words, you need something like $\displaystyle f(x_0)=g(x_0)$, which is true, but why?

6. Originally Posted by tn11631
Suppose f: [a,b] -> R and g: [a,b]-> R are both continuous. Let T={x: f(x)=g(x)}. Prove that T is closed.
More generally the follow works whenever the codomain is Hausdorff.

Consider that $\displaystyle A(f,g)=\left\{x\in[a,b]:f(x)=g(x)\right\}=\varphi^{-1}(\Delta)$ where $\displaystyle \Delta$ is the diagonal and $\displaystyle \varphi:[a,b]\mapsto\mathbb{R}\times\mathbb{R}$ given by $\displaystyle x\mapsto(f(x),g(x))$

7. T is the inverse image of {0} (a closed set in R) under f-g (a continuous map).

8. As a note of interest this shows that if $\displaystyle A(f,g)$ contains a dense subset of $\displaystyle \text{Dom }f$ then $\displaystyle f=g$