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Let $f: \mathbb R \to \mathbb R $ and $g: \mathbb R \to \mathbb R$ be continuous functions. Show the set $ E = \{ x \in\mathbb R: f(x)=g(x)\} $ is closed.

My approach

A solution I found is the following:

$h=f-g$

$h(x)=f(x)-g(x)=0$

f and g are continious and h is continious

Taking $h(E)=\{ x\in \mathbb R:h(x)=0\}$ // makes no sense, why $x\in\mathbb R$ and not $h(x)\in\mathbb R$

$h^{-1} (E)=\{0\}$ is closed // why is the inverse only zero?

=> $E$ is closed

Is the solution correct? Seems very elegant and short, but $h(E)$ makes no sense to me please explain.

5xum
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user2253741
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1 Answers1

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How much do you know about continuous functions? For example, do you know that for a continuous function $h$ and a closed set $X$, the set $h^{-1}(X)$ is always closed?

If you know that, the continuation is badly written, but captures the idea. First, you define the function $h = f-g$. Then, you can see that $$E=\{x\in \mathbb R: f(x) = g(x)\} = \{x\in\mathbb R: f(x)-g(x)=0\} = \{x\in\mathbb R: h(x) = 0\} = h^{-1}(\{0\})$$

This means that $E$ is the preimage of $\{0\}$ for the function $h$, and, since $h$ is continuous and $\{0\}$ is a closed set, $E$ is a closed set.

5xum
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  • it is known that f is cont E subset of X is closed than f^-1 (E) is closed. – user2253741 Jun 16 '14 at 06:59
  • @user2253741 If you know that, than the second part of my question should explain everything. – 5xum Jun 16 '14 at 07:01
  • can you please answer the 2 questions I asked explicitly. I still have hard time understanding. – user2253741 Jun 16 '14 at 07:21
  • @user2253741 Rather than understanding what you wrote (which is at best confuzing and actually quite wrong, since $h^{-1}(E)$ is not ${0}$), take a look at my answer (which is probably what your answer tried, but failed to, be. – 5xum Jun 16 '14 at 07:24
  • can you explain what is the deal with the inverse of the function of the set? I find it hard to understand. h−1({0}), also what is the deal with the {0} set I didn't understand your explanation maybe you could use some reference. – user2253741 Jun 16 '14 at 07:35
  • $h^{-1}(A)$ is not the inverse function of $h$, it is the preimage of the set $A$. By definition, $h^{-1}(A) = {x\in X, h(x) \in A}$. – 5xum Jun 16 '14 at 07:38