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Let $c$ and $s$ be two real continuous functions on $\mathbb{R}$ and $P= \left \{ x \in \mathbb{R} : c(x) > s(x) \right \}$. Prove that $P$ is open.

This is what I'm thinking...

Let $P= \left \{ x \in \mathbb{R} : c(x) > s(x) \right \}$be given. Since $c$ and $s$ are continuous on $\mathbb{R}$, then $c-s$ is also continuous on $\mathbb{R}$. Hence, $P= \left \{ x \in \mathbb{R} : (c-s)(x) > 0 \right \}$.

Case 1: Let $(c-s)(x) \leq 0$, $\forall x \in \mathbb{R}$. Then $P = \emptyset$ and thus $P$ is an open set.

Case 2: Let $(c-s)(x) > 0$, $\forall x \in \mathbb{R}$. Then $P = \mathbb{R}$ and thus $P$ is an open set.

Case 3: Let $P$ be a proper subset of $\mathbb{R}$. Let $t \in P$. Then $(c-s)(t) >0$ since $(c-s)$ is continuous on $\mathbb{R}$ and $(c-s)(t) >0$, then by the neighborhood property there exists a $\epsilon > 0$ such that $\forall x \in N_\epsilon(t)$, $(c-s)(x) > 0$. Therefore, $N_\epsilon (t) \subset P$. Thus, if $t \in P$ then $N_\epsilon (t) \subset P$. Therefore $t$ is an interior point of $P$. Since $t$ is arbitrary, every point of $P$ is an interior point of $P$ and hence $P$ is an open set.

Is this how you would prove this?

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  • $\begingroup$ Note that the pre-image of an open set under a continuous function is open!! (By definition of a continuous function). All we want is the pre-image of $(0, \infty)$ under continuous $(c-s)$. $\endgroup$ – SEWillB May 21 '17 at 20:09
  • $\begingroup$ So if the proof were to now read... Let $P= \left \{ x \in \mathbb{R} : c(x) > s(x) \right \}$be given. Since $c$ and $s$ are continuous on $\mathbb{R}$, then $c-s$ is also continuous on $\mathbb{R}$. Hence, $P= \left \{ x \in \mathbb{R} : c(x) > s(x) \right \} = \left \{ x \in \mathbb{R} : (c-s)(x) > 0 \right \} = (c-s)^{-1} (0, \infty)$. Since, $(c-s): \mathbb{R} \rightarrow \mathbb{R}$ is continuous then the inverse image of every open set is open. Since $(0, \infty)$ is open in $\mathbb{R}$ then $(c-s)^{-1} (0, \infty)$ is open. Therefore, $P$ is open. Would this be the correct proof? $\endgroup$ – user6259845 May 22 '17 at 0:32
  • $\begingroup$ What do you mean by "the neighborhood property"? Which definition of a continuous function do you use (or are expected to use)? If I were to edit your approach, I would discard cases 1,2, only do case 3, and when you discuss $(c-s)(t)>0$ it may help to fix $d=\frac{(c-s)(t)}2>0$ and do something with it. $\endgroup$ – Mirko May 22 '17 at 0:36
  • $\begingroup$ That works perfectly! $\endgroup$ – SEWillB May 22 '17 at 0:43
  • $\begingroup$ @Mirko, here is the definition we use...Let $E$ be a subset of $\mathbb{R}$ and $f$ a real-valued function with domain $E$. The function $f$ is continuous at a point $ p \in E$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|f(x)-f(p)| < \epsilon$ for all $x \in E$ with $|x-p| < \delta$. The function f is continuous on E if and only if f is continuous at every point $p \in E$. $\endgroup$ – user6259845 May 22 '17 at 0:45
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You could come up with different answers, depending on which definition of continuity you use.

Let me use the definition provided in your comment (which I edited a bit for the case $E=\mathbb R$ and $p=t$).

Let $f$ be a real-valued function with domain $\mathbb R$. The function $f$ is continuous at a point $t\in\mathbb R$ if for every $\varepsilon>0$ , there exists a $\delta>0$ such that $|f(x)−f(t)|<\varepsilon$ for all $x$ with $|x−t|<\delta$, that is, for all $x\in N_\delta(t):=(t-\delta,t+\delta)$. The function $f$ is continuous on $\mathbb R$ if and only if f is continuous at every point $t\in\mathbb R$.

Your proof is correct, but you may discard cases 1 and 2, as they are redundant. In case 3, the phrase "by the neighborhood property" may be a bit vague (what is "the neighborhood property" ... perhaps it is defined in your book, but it is not such a basic definition as that of continuity, and needs either to be stated along with your proof, or to be avoided and replaced with a self-contained detailed argument). Here is how I would edit your original proof.

Let $P= \left \{ x \in \mathbb{R} : c(x) > s(x) \right \}$. To show that $P$ is open, it is enough to show that if $t\in P$
then $t$ is an interior point of $P$ (that is, there is a neighborhood of $t$ entirely contained in $P$).
[Note that I need not consider cases, like $P=\emptyset$, or $P=\mathbb R$, and I do not claim that there are $t\in P$, but only that, IF $\ t\in P\ $ THEN it must be an interior point.]
Continuing with your proof, suppose $t\in P$ for some $t$. Then $(c−s)(t)>0$. Say $(c−s)(t)=a>0$. Let $\varepsilon=\frac a2$, then clearly $\varepsilon>0$. Since $(c−s)$ is continuous on $\mathbb R$ and $(c−s)(t)=a$, there exists a $\delta>0$ such that $\forall x\in N_\delta(t)$ we have that $|(c−s)(x)-(c−s)(t)|= |(c−s)(x)-a|<\frac a2=\varepsilon$. Thus $\frac {-a}2<(c−s)(x)-a<\frac a2$, and $\frac a2<(c−s)(x)<\frac{3a}2$. In particular, $0<\frac a2<(c−s)(x)$, $\forall x\in N_\delta(t)$, therefore, $N_\delta(t)\subseteq P$. Thus, if $t\in P$ then $N_\delta(t)\subseteq P$ for some $\delta>0$. Therefore $t$ is an interior point of $P$. Since $t$ was arbitrary, every point of $P$ is an interior point of $P$, and hence $P$ is an open set.

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You can do it much easier than that. Since $c$ and $s$ are continuous, so is $c-s$. Then you can write $P = \{x \in \mathbb{R} | (c-s)(x)>0\}$. You know that the preimage of an open set under a continuous mapping is open, so $P$ is open.

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Let $f = c - s$. Then $$\{x : c(x) - s(x) > 0\} = f^{-1}((0,\infty)).$$

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