Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be a real-valued function on $\mathbb{R}$. Show $f$ is continuous if and only if $f$ is both upper and lower semi-continuous, using the definition of continuity and semi-continuity based on open set.

It's difficult for me to be rigourous here. For example for necessity. Let $O=\{f(x) : f(x) > a, x \in \mathbb{R}\}$ with $a$ real. By continuity, if $O$ is open then the pre-image of $O$ is open whence $f$ is lower semi-continuous. Is $O$ open? If so why? This point blocks me.

I think I can do sufficiency correctly though. Suppose $f$ is both upper and lower semi-continuous. Let $O$ be an open set.  Since any open set in $\mathbb{R}$ can be expressed as a union of open sets of the form $(a,b)$, then $O=\cup (a_n,b_n)$ for some open sets $(a_n,b_n)$. Since each $(a_n,b_n)$ can be written has $(a_n,\infty) \cap (-\infty,b_n)$, by semi-continuity, the pre-image of each $(a_n,b_n)$ is open. Since $f^{-1}(O)=f^{-1}(\cup(a_n,b_n))=\cup f^{-1}(a_n,b_n)$, then the pre-image of $O$ is open anf $f$ is continuous.

Definition: A function is continuous if and only if the pre-image of any open set is open. A function is lower semi-continuous if and only if the set $\{x \in \mathbb{R} : f(x) > a\}$ is open for any $a$ real. A function is upper semi-continuous if and only if the set $\{x \in \mathbb{R} : f(x) < a\}$ is open for any $a$ real.

share|improve this question
    
Using the open set definition is possible, but is useless pain. See my answer. If you absolutely wanted to use the open set definition, I must say I've never seen a semi-continuity definition based on open sets ; please put yours here. –  Patrick Da Silva Feb 17 '12 at 22:39
1  
To summarize Patrick's answer: your proof of sufficiency is essentially correct (although it seems you're mixing up "image" and "preimage" at one point). For necessity, your definition of $O$ is wrong, and it seems to be the main problem. Look up Patrick's solution if you want to know what went wrong. –  M Turgeon Feb 17 '12 at 23:17
    
@MTurgeon Thanks for pointing the error in my sufficiency proof. I corrected it. –  Nicolas Essis-Breton Feb 18 '12 at 23:13
add comment

1 Answer

up vote 1 down vote accepted

What you want is use the $\varepsilon$-$\delta$ definition of continuity/semi-continuity

$$ \forall \varepsilon > 0, \quad \exists \delta > 0 \quad \text{s.t.} \quad |x-x_0| < \delta \quad \Longleftrightarrow \quad |f(x) - f(x_0)| < \varepsilon. $$ Since $|f(x) - f(x_0)| < \varepsilon$ is equivalent to $f(x_0) - \varepsilon < f(x) < f(x_0) + \varepsilon$, this statement is equivalent to $$ \forall \varepsilon > 0, \quad \exists \delta > 0 \quad \text{s.t.} \quad |x-x_0| < \delta \quad \Longleftrightarrow \quad f(x_0) - \varepsilon < f(x) < f(x_0) + \varepsilon $$ which means continuity is equivalent to upper and lower semi-continuity.

ADDED : I'll assume the definition of semi-continuity is the following, i.e. that the preimage of an open set of the form $\{ y \, | \, y > a \}$ is open for any upper semi-continuous function, and that the preimage of an open set of the form $\{ y \, | \, y < a \}$ is open for any lower semi-continuous function. If this is not the definition you have, let me know. I'm just guessing those definitions from the $\varepsilon$-$\delta$ definitions of continuity.

We show that continuity $\Longleftrightarrow$ upper and lower semi-continuity.

$(\Longrightarrow)$ This one is clear, since if the pre-image of any open set is open, then in particular are the pre-images of those of the form $(a, \infty)$ and $(-\infty,a)$.

$(\Longleftarrow)$ If $f^{-1}((a,\infty))$ and $f^{-1}((-\infty,b))$ are open sets, then since $f^{-1}((a,b)) = f^{-1}((a,\infty)) \cap f^{-1}((-\infty,b))$ and that the intersection of two open sets is open, then $f^{-1}((a,b))$ is open. Now any open set $\mathcal O$ in $\mathbb R$ is of the form $$ \bigcup_{n=0}^{\infty} (a_n, b_n) $$ where all the intervals $(a_n,b_n)$ are pairwise disjoint and $a_n \in \mathbb R \cup \{-\infty\}$, $b_n \in \mathbb R \cup \{\infty\}$. But $$ f^{-1} \left( \bigcup_{i \in I} \mathcal O_i\right) = \bigcup_{i \in I} f^{-1} (\mathcal O_i) $$ for any collection of sets indexed by any set $I$ (prove this trivially by the definition of pre-images and show $\subseteq$/$\supseteq$), thus, $$ f^{-1}(\mathcal O) = \bigcup_{n=0}^{\infty} f^{-1}((a_n,b_n)) $$ which is an open set.

Hope that helps,

share|improve this answer
    
Like I always say : downvoters, your downvotes are pointless if you don't justify yourself. Do it here please. –  Patrick Da Silva Feb 17 '12 at 22:59
    
That's why I think votes shouldn't be anonymous. –  Pedro Tamaroff Feb 17 '12 at 23:06
1  
@Peter : Exactly, I agree. This way downvoters would feel the need to explain themselves, otherwise they would only attract hate towards them by downvoting. –  Patrick Da Silva Feb 17 '12 at 23:16
1  
Your definition is not different. $\{ x \in \mathbb R \, | \, f(x) > a \} = \{ x \in \mathbb R \, | \, f(x) \in (a, \infty) \} = f^{-1} ((a,\infty))$. Is that okay? (The same goes for lower semi-continuity.) –  Patrick Da Silva Feb 17 '12 at 23:45
1  
Thank you Patrick, all the concepts are link in my head now. It's great. –  Nicolas Essis-Breton Feb 17 '12 at 23:49
show 8 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.