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Lef a function $f$ be defined and continuous on $R^n$, the range of $f$ is the extended real numbers. My book claims that the set $\{x\in R^n :f(x)>a\}$ is open for all $a \ \in R^n$. I am wondering why this is the case?

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2 Answers 2

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Depends on how you define continuous. If you define a continuous function as

$f: X\to Y$ is continuous if and only if, for every open set $U\subseteq Y$, the set $f^{-1}(U)$ is an open set.

then the statement your book makes is almost trivial (you just need to prove that the set $\{y\in\mathbb R| y>a\}$ is an open set)


However, you may be operating with the more standard definition of continuity:

$f:\mathbb R^n\to\mathbb R$ is continuous if and only if, for every $x_0\in\mathbb R^n$ and every $\epsilon>0$, there exists some $\delta>0$ such that for every $x$ for which $\|x-x_0\|<\delta$, we also have $|f(x)-f(x_0)|<\epsilon$.

In which case the proof is slightly longer, but still not complicated.

You need to take an arbitrary $x_0\in U$ (where $U=\{x\in\mathbb R^n|f(x)>a\}$), and find some $\epsilon >0$ such that $$\{x|\|x-x_0\|<\epsilon\}\subseteq U$$

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  • $\begingroup$ How do we define a open set in the extended real numbers? $\endgroup$
    – Allitee
    Mar 24, 2016 at 14:10
  • $\begingroup$ @Allitee Sorry to say this, but if you don't know the definition of an open set, then the question you are asking is currently too advanced for you. I suggest you either go back a couple of chapters or start with an easier textbook. $\endgroup$
    – 5xum
    Mar 24, 2016 at 14:13
  • $\begingroup$ My book defines a set O is open in $R^n$ if for every point is contain in some open ball $B\subset O$. But it doesn't mention the open set in the extended real numbers. In your second explanation, can $f(x)=\infty \ for \ some \ x$? $\endgroup$
    – Allitee
    Mar 24, 2016 at 14:20
  • $\begingroup$ @Allitee Yes, and the standard topology on the extended reals (which is, I have no doubt about it, also defined in your book) is that $(a,\infty]$ is also an open set. $\endgroup$
    – 5xum
    Mar 24, 2016 at 14:24
  • $\begingroup$ So if I take $x_0\in U$, and $f(x_0)=\infty$, how do I show that $ \exists r>0$ such that $\{ x : ||x-x_0||<r \} \subset U $? $\endgroup$
    – Allitee
    Mar 24, 2016 at 14:33
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Call your set $L_a$, and fix $x_0\in L_a$. To prove $L_a$ is open, we need to find an open ball $B_\delta(x_0)$ around $x_0$ so $B_\delta(x_0)\subset L_a$. In other words, we need to show that if $x$ stays close enough to $x_0$, all values $f(x)$ will be larger than $a$.

First assume $f(x_0)$ isn't infinite. Define $\epsilon$ to be $f(x_0)-a$. Then, because $f$ is continuous, we know we can choose a $\delta$ so that if $x$ is in the $\delta$-ball centered at $x_0$, then $f(x)$ is in the $\epsilon$-ball centered at $f(x_0)$. In other words, we can choose $\delta$ so $$x\in B_\delta(x_0)\implies|f(x)-f(x_0)|<\epsilon=f(x_0)-a$$ But this means $a-f(x_0)<f(x)-f(x_0)<f(x_0)-a$. The left inequality implies $a<f(x)$.

What about the case where $f(x_0)$ is infinite? Well, the open sets that contain infinity all contain a set of the form $(y,\infty]$. So in this case, continuity at $x_0$ means that for any $y\in\mathbb{R}$ we can find a $\delta$ so that $$x\in B_\delta(x_0)\implies f(x)\in(y,\infty]$$ Choosing $y=a$ gives the result.

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