# Characterizing continual features based upon the chart of the function

I had actually asked this inquiry : Characterising Continuous functions time back, and also this inquiry is essentially pertaining to that inquiry.

Intend we have a function $f: \mathbb{R} \to \mathbb{R}$ and also intend the set $G = \\{ (x,f(x) : x \in \mathbb{R}\\}$ is attached and also enclosed $\mathbb{R}^{2}$, after that does it indicate $f$ is continual?

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2019-05-13 03:52:19
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intend you have a function $f:\mathbb{R} \rightarrow \mathbb{R}$ which is not continual and also its chart is shut and also attached. wlog think f is not continual at absolutely no and also f (0) = 0.

first, I intend to show that when x comes close to absolutely no after that either f (x) comes close to 0 or mosts likely to infinity. for every single b > a > 0 allowed $X_{a,b} = \{ x>0 \mid a < f(x) < b \}$. if this set is not vacant, after that its infimum can not be 0, due to the fact that after that the closure of the chart will certainly offer us that $a \leq f(0) \leq b$ (you can locate a collection $a_n \rightarrow 0 \subseteq X_{a,b}$ and also a below - collection such that $f(a_{n_i})$ is monotone and also hence merges in $R^2$).

currently I intend to make use of the connectedness to show that f (x) can not most likely to infinity when x mosts likely to absolutely no.

$f$ isn't continual at absolutely no, so wlog there is an $\varepsilon > 0$ such that $X_{2\varepsilon, \infty}$ is not vacant and also its infimum is 0. $X_{\varepsilon, 2\varepsilon}$ has infimum > 0 (otherwise vacant) - represent it by $\delta$. if you have $0 < x_0 < \delta$ such that $x_0 \notin X_{2 \varepsilon, \infty}$ (therefore $f(x_0) < \varepsilon$) after that you have the open collections $A = (0, x_0) \times (1.5 \varepsilon, \infty)$ and also $R^2 - \bar A$ which divides the garph of the function.

or else, if $(0,\delta) \subseteq X_{2\varepsilon, \infty}$ after that you can take the collections $A = \{x\leq 0 \} \cup (0, \delta/2) \times (-\infty, 1.5 \varepsilon)$ and also $R^2 - \bar A$.

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2019-05-17 15:04:31
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The solution is of course. Below is one means to confirm it. (There could be a slicker means, yet this appears to function.)

Think $G$ is attached and also shut. Allow $a\in\mathbb R$ be approximate, and also allow $\epsilon>0$ be offered. Due to the fact that $(a,f(a)-\epsilon)\notin G$, the reality that the enhance of $G$ is open indicates that there is an item area of the kind $(a-\delta,a+\delta)\times (f(a)-\epsilon-c, f(a)-\epsilon+c)$ had in the enhance of $G$. This suggests that $|x-a|<\delta$ indicates that of the adhering to 2 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 kind of $x\in (a-\delta,a+\delta)$ to make sure that the 2nd inequality holds, claim $f(b)\le f(a)-\epsilon-c$ (without loss of generalization, we might think $b\lt a$), after that the chart of $f$ does not converge the adhering to 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 layout listed below.) This busted line separates $G$, negating the presumption that $G$ is attached. Consequently inequality (1) holds when $|x-a|<\delta$.
A comparable argument reveals that $f(x)\lt f(a)+\epsilon$ when $|x-a|<\delta$.

Placing these with each other, we end that $|x-a|<\delta$ indicates $|f(x)-f(a)|<\epsilon$, so $f$ is continual at $a$. 0
2019-05-17 15:03:54
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Yes, I assume so.

First, observe that such $f$ has to have the intermediate value building. For intend not ; after that there exist $a < b$ with (say) $f(a) < f(b)$ and also $y \in (f(a),f(b))$ such that $f(x) \ne y$ for all $x \in (a,b)$. After that $A = (-\infty,a) \times \mathbb{R} \cup (-\infty,b) \times (-\infty,y)$ and also $B = (b, +\infty) \times \mathbb{R} \cup (a,+\infty) \times (y,+\infty)$ are disjoint nonempty open parts of $\mathbb{R}^2$ whose union has $G$, negating connectedness. (Draw an image.)

Currently take some $x \in \mathbb{R}$, and also intend $f(x) < y < \limsup_{t \uparrow x} f(t) \le +\infty$. After that there is a series $t_n \uparrow x$ with $f(t_n) > y$ for each and every $n$. By the intermediate value building, for each and every $n$ there is $s_n \in (t_n, x)$ with $f(s_n) = y$. So $(s_n, y) \in G$ and also $(s_n,y) \to (x,y)$, so given that $G$ is shut $(x,y) \in G$ and also $y = f(x)$, an opposition. So $\limsup_{t \uparrow x} f(t) \le f(x)$. In a similar way, $\limsup_{t \downarrow x} f(t) \le f(x)$, so $\limsup_{t \to x} f(t) \le f(x)$. In a similar way, $\liminf_{t \to x} f(t) \ge f(x)$, to make sure that $\lim_{t \to x} f(t) = f(x)$, and also $f$ is continual at $x$.

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2019-05-17 15:02:01
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This Monthly paper has brief straightforward evidence of the adhering to

THEOREM $\$ TFAE if $\rm\ f: \mathbb R\to \mathbb R\$ has a shut chart in $\:\mathbb R^2$
(a) $\rm\ \ f\$ is continual.
(b) $\rm\ \ f\$ is in your area bounded.
(c) $\rm\ \ f\$ has the intermediate value building.
(d) $\rm\ \ f\$ has a linked chart in $\rm\mathbb R^2$.

Extra usually the outcome is just a grandfather clause of R. L. Moore is 1920 characterization of a topological line as an in your area portable statistics room that is divided right into 2 linked collections by each of its factors.

Per demand, I've added the evidence of the theory listed below. . 0
2019-05-17 14:42:01
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