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Let $(M,d)$ be a metric space and $f,g:(M,d) \rightarrow \mathbb{R}$ be continuous functions. Show the set $A=\{x \in M: f(x)<g(x)\}$ is an open set in $(M,d)$.

To show the statement we need to take a point $x_0$ and show that there exist a ball around $x_0$ which is contained in $A$. My proof is the following but it does not reasonable to me that I have to have the condition $\epsilon_1>\epsilon_2$.

Let $x_0 \in A \rightarrow f(x_0)<g(x_0)$

Since $f$ is continuous $\forall \epsilon_1>0\,\,\exists\,\,\,\delta_1>0\,\,\,\text{s.t.} \,\,\,d(x,x_0)<\delta_1 \,\,\,\rightarrow |f(x)-f(x_0)|<\epsilon_1$ and $$ -\epsilon_1<f(x)-f(x_0)<\epsilon_1 \,\, \Rightarrow f(x)-f(x_0)<\epsilon_1 \tag{1} $$

And since $g$ is continuous $\forall \epsilon_2>0\,\,\exists\,\,\,\delta_2>0\,\,\,\text{s.t.} \,\,\,d(x,x_0)<\delta_2 \,\,\,\rightarrow |g(x)-g(x_0)|<\epsilon_2$ and $$ -\epsilon_2<g(x)-g(x_0)<\epsilon_2 \,\, \Rightarrow -\epsilon_2 <g(x)-g(x_0)\tag{2} $$

let $\delta=\min (\delta_1,\delta_2)$ so using (1) and (2) $$ f(x)-f(x_0)-\epsilon_2<\epsilon_1 +g(x)-g(x_0) \Rightarrow -\epsilon_2+\epsilon_1< g(x)-f(x) $$ Since $f(x_0)-g(x_0)<0$.

Therefore, if $ \epsilon_1>\epsilon_2 \Rightarrow 0< g(x)-f(x)$ so $d(x,x_0)<\delta$ is contained in $A$.

Could you help me to get rid of this confusion. I understand that because $f,g$ are continuous those statement should be held for all $\epsilon$ but having this condition does not make sense to me.

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  • $\begingroup$ .... so using (1) and (2): how do you infer the next inequality (at that point in your text)? I do not see it. $\endgroup$
    – Jimmy R.
    Nov 5 '18 at 6:24
  • $\begingroup$ I summed over (1) and (2). $\endgroup$
    – Saeed
    Nov 5 '18 at 16:48
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Your set is $$(g-f)^{-1}(]0,\infty[)$$

This is open, as inverse image of an open set under a continuous function.

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Since you're not proving continuity, but using continuity, you want to pick a specific $\epsilon$.

Case in point: if $f(x)=0, g(x)=x, x_0=1$ and $\epsilon=2$, then your $\delta$ becomes $2$, and $(-1,3)$ is certainly not a subset of your set. So an arbitrary $\epsilon$ isn't going to do you any good.

I claim that $\epsilon=\frac{g(x_0)-f(x_0)}2$ works. Let $\delta_f>0$ be such that for any $x_1\in M$ with $d(x_0,x_1)<\delta_f$, we have $|f(x_0)-f(x_1))|<\epsilon$, and let $\delta_g$ be defined analogously. Pick $\delta=\min\{\delta_f,\delta_g\}$. Then, for any $x_1\in M$ with $d(x_0,x_1)<\delta$, we have $$ f(x_0)+\frac{g(x_0)-f(x_0)}2=g(x_0)-\frac{g(x_0)-f(x_0)}2\\ f(x_0)+\epsilon=g(x_0)-\epsilon\\ f(x_1)<f(x_0)+\epsilon=g(x_0)-\epsilon<g(x_1) $$ where the first inequality is because $d(x_0,x_1)<\delta_f$ and the second one because $f(x_0,x_1)<\delta_g$.

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  • $\begingroup$ The result is true for any topological space $M$. $\endgroup$ Nov 5 '18 at 8:39
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    $\begingroup$ @DanielWainfleet And the proof is much easier in a topological context. But, alas, we are given a metric space and the $\epsilon$-$\delta$ definition of continuity. $\endgroup$
    – Arthur
    Nov 5 '18 at 8:52
  • $\begingroup$ @Arthur Actually, can we look at the proof if M was a generic topological space? I was thinking if $f$ and $g$ are continuous functions, $h(x) = (f - g)(x)$ is a continuous function. If $x_0 \in M$, and if I were to define an open neighborhood of $x_0$, $N(x,\epsilon)$, could we say that $h(N(x,\epsilon))$ is an open set that contains the interior point $x_0$? $\endgroup$ Jan 30 '20 at 5:01
  • $\begingroup$ @Overachiever If $M$ were a generic topological space, we would have to discard the idea of $\varepsilon$-$\delta$ altogether. In that case, the function $h$ is continuous, so by definition of continuous (inverse image of any open is open), the set of points in $M$ where $h$ is strictly negative is open. So it's actually a lot easier. See the other answer. $\endgroup$
    – Arthur
    Jan 30 '20 at 7:10

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