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I had asked this question: http://math.stackexchange.com/questions/4412/characterising-continuous-functions some time back, and this question is more or less related to that question.

Suppose we have a function $f: \mathbb{R} \to \mathbb{R}$ and suppose the set $G = \{ (x,f(x) : x \in \mathbb{R}\}$ is connected and closed in $\mathbb{R}^{2}$, then does it imply $f$ is continuous?

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4 Answers 4

up vote 12 down vote accepted

This Monthly paper has short simple proofs of the following

THEOREM$\ $ TFAE if $\rm\ f: \mathbb R\to \mathbb R\ $ has a closed graph in $\:\mathbb R^2$
(a)$\rm\ \ f\ $ is continuous.
(b)$\rm\ \ f\ $ is locally bounded.
(c)$\rm\ \ f\ $ has the intermediate value property.
(d)$\rm\ \ f\ $ has a connected graph in $\rm\mathbb R^2$.

More generally the result is merely a special case of R. L. Moore's 1920 characterization of a topological line as a locally compact metric space that is separated into two connected sets by each of its points.

Per request, I've appended the proof of the theorem below.

alt text alt text

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@Bill: Thanks a lot! –  anonymous Sep 30 '10 at 19:06
    
@Bill Dubuque It is a pity that some of us don't have access to the paper, maybe you could add a screen shot? –  AD. Sep 30 '10 at 21:13
3  
@AD.: Done, enjoy! –  Bill Dubuque Sep 30 '10 at 21:36
    
@Bill Dubuque: Thanks! –  AD. Oct 1 '10 at 5:07
    
This answer triggered a discussion on meta. See meta.math.stackexchange.com/questions/529/…. –  KennyTM Oct 6 '10 at 9:33

The answer is yes. Here's one way to prove it. (There might be a slicker way, but this seems to work.)

Assume $G$ is connected and closed. Let $a\in\mathbb R$ be arbitrary, and let $\epsilon>0$ be given. Because $(a,f(a)-\epsilon)\notin G$, the fact that the complement of $G$ is open implies that there is a product neighborhood of the form $(a-\delta,a+\delta)\times (f(a)-\epsilon-c, f(a)-\epsilon+c)$ contained in the complement of $G$. This means that $|x-a|<\delta$ implies that one of the following two inequalities holds:

  1. $f(x)\ge f(a)-\epsilon+c>f(a)-\epsilon$, or
  2. $f(x)\le f(a)-\epsilon-c$.

If there is any $x\in (a-\delta,a+\delta)$ so that the second inequality holds, say $f(b)\le f(a)-\epsilon-c$ (without loss of generality, we may assume $b\lt a$), then the graph of $f$ does not intersect the following set:

$$\{(b,y): y\ge f(a)-\epsilon \}\cup \{(x,f(a)-\epsilon): b\le x \le a\} \cup \{(a,y): y\le f(a)-\epsilon\}.$$

(See the diagram below.) This broken line disconnects $G$, contradicting the assumption that $G$ is connected. Therefore inequality (1) holds when $|x-a|<\delta$.
A similar argument shows that $f(x)\lt f(a)+\epsilon$ when $|x-a|<\delta$.

Putting these together, we conclude that $|x-a|<\delta$ implies $|f(x)-f(a)|<\epsilon$, so $f$ is continuous at $a$.

alt text

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suppose you have a function $f:\mathbb{R} \rightarrow \mathbb{R}$ which is not continuous and its graph is closed and connected. wlog assume f is not continuous at zero and f(0)=0.

first, I want to show that when x approaches zero then either f(x) approaches 0 or goes to infinity. for every b>a>0 let $ X_{a,b} = \{ x>0 \mid a < f(x) < b \} $ . if this set is not empty, then its infimum cannot be 0, because then the closure of the graph will give us that $ a \leq f(0) \leq b $ (you can find a series $a_n \rightarrow 0 \subseteq X_{a,b}$ and a sub-series such that $f(a_{n_i})$ is monotone and thus converges in $R^2$).

now I want to use the connectedness to show that f(x) cannot go to infinity when x goes to zero.

$f$ isn't continuous at zero, so wlog there is an $\varepsilon > 0$ such that $ X_{2\varepsilon, \infty}$ is not empty and its infimum is 0. $X_{\varepsilon, 2\varepsilon}$ has infimum > 0 (if not empty) - denote it by $\delta$. if you have $0 < x_0 < \delta$ such that $ x_0 \notin X_{2 \varepsilon, \infty} $ (and so $f(x_0) < \varepsilon $ ) then you have the open sets $ A = (0, x_0) \times (1.5 \varepsilon, \infty) $ and $ R^2 - \bar A$ which separates the garph of the function.

otherwise, if $ (0,\delta) \subseteq X_{2\varepsilon, \infty} $ then you can take the sets $A = \{x\leq 0 \} \cup (0, \delta/2) \times (-\infty, 1.5 \varepsilon)$ and $R^2 - \bar A$.

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Yes, I think so.

First, observe that such $f$ must have the intermediate value property. For suppose not; then there exist $a < b$ with (say) $f(a) < f(b)$ and $y \in (f(a),f(b))$ such that $f(x) \ne y$ for all $x \in (a,b)$. Then $A = (-\infty,a) \times \mathbb{R} \cup (-\infty,b) \times (-\infty,y)$ and $B = (b, +\infty) \times \mathbb{R} \cup (a,+\infty) \times (y,+\infty)$ are disjoint nonempty open subsets of $\mathbb{R}^2$ whose union contains $G$, contradicting connectedness. (Draw a picture.)

Now take some $x \in \mathbb{R}$, and suppose $f(x) < y < \limsup_{t \uparrow x} f(t) \le +\infty$. Then there is a sequence $t_n \uparrow x$ with $f(t_n) > y$ for each $n$. By the intermediate value property, for each $n$ there is $s_n \in (t_n, x)$ with $f(s_n) = y$. So $(s_n, y) \in G$ and $(s_n,y) \to (x,y)$, so since $G$ is closed $(x,y) \in G$ and $y = f(x)$, a contradiction. So $\limsup_{t \uparrow x} f(t) \le f(x)$. Similarly, $\limsup_{t \downarrow x} f(t) \le f(x)$, so $\limsup_{t \to x} f(t) \le f(x)$. Similarly, $\liminf_{t \to x} f(t) \ge f(x)$, so that $\lim_{t \to x} f(t) = f(x)$, and $f$ is continuous at $x$.

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See also exercise 4.19 in Baby Rudin. –  Nate Eldredge Sep 30 '10 at 16:11
    
... and references therein. –  Nate Eldredge Sep 30 '10 at 16:16

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