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Assume the function $h\colon \Bbb R\to \Bbb R$ with dense graph such that $h\restriction C$ connected for any connected set $C\subset\Bbb R.$ let $f\colon (0,1)\to \Bbb R$ be a homeomorphism.

Now define a function $g\colon \Bbb R\to \Bbb R$ as $g=h\circ f$ on $(0,1)$ and 0 on its complement. I want to check if the graph of $g$ will be connected or not. It is clearly that the graph will be as union of three connected pieces.

I need to check if the whole graph will be connected or not. I think it will be problem in the end points. Any help will be helpful.

00GB
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    Is $h(x)=x$ a counterexample, or am I misunderstanding the requirement on $h$? – Karl Mar 14 '21 at 16:03
  • @Karl, $h$ must have dense graph – 00GB Mar 14 '21 at 16:06
  • Dense in $\Bbb R^2$, right? How does this correspond to your condition involving $C$? The graph of $h\restriction C$ is indeed connected for any connected set $C\subset \Bbb R$, for $h(x)=x$. – Karl Mar 14 '21 at 16:13
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    Maybe you mean ${(x,h(x)):x\in\Bbb R}\cap C$ is connected for any connected $C\in\Bbb R^2$? – Karl Mar 14 '21 at 16:16
  • @Karl, By $h\restriction C={(x,h(x))\colon x\in C}$ . – 00GB Mar 14 '21 at 16:19
  • Then $h(x)=x$ works because ${(x,x):x\in C}$ is connected if $C$ is connected. – Karl Mar 14 '21 at 16:20
  • @Karl, $h$ must have a dense graph in $\Bbb R^2$ – 00GB Mar 14 '21 at 16:24
  • Oh I see, these are two separate conditions. Do we know such an $h$ exists? – Karl Mar 14 '21 at 16:28
  • If I do not misunderstand something, the graph of $h$ is connected and locally connected. But then $h$ is continuous (see https://math.stackexchange.com/q/4044406). Hence the graph of $h$ cannot be dense. – Paul Frost Mar 15 '21 at 00:51
  • @PaulFrost, why $h$ is ;locally connected – 00GB Mar 15 '21 at 01:20
  • Brian Scott. Do you have any idea ? – 00GB Mar 15 '21 at 02:51
  • @00GB If $C$ is an open interval, then $h\restriction C$ is open in the graph because it is the intersection of the graph and the strip $C \times \mathbb R$. Hence the graph has a basis consisting of connected open sets. – Paul Frost Mar 15 '21 at 08:35
  • @PaulFrost, there is something wrong the function $h$ with this property it might be discontinuous at any point. $h\restriction C={(x,h(x))\colon x\in\Bbb R}$ – 00GB Mar 15 '21 at 13:14
  • @00GB It seems that you have doubts. Is it about the fact that the graph is locally connected? If you accept that it is true, then https://math.stackexchange.com/q/4044406 applies to show that $h$ is continuous. – Paul Frost Mar 16 '21 at 00:57
  • @PaulFrost, It is about the $h$ has locally connected graph – 00GB Mar 16 '21 at 01:17
  • @PaulFrost, Could you explain based in my definition of $h$ how you conclude the graph of $h$ is locally connected ? – 00GB Mar 18 '21 at 03:21
  • @00GB I was wrong. My error was this: For $x_0 \in \mathbb R$ let $U_n(x_0) = (x_0-1/n,x_0+1/n)$. Then the $V_n = h \restriction U_n(x_0)$ are open connected neigborhoods of $(x_0,h(x_0))$ such that $\overline{V_{n+1}} \subset V_n$ and $\bigcap V_n = {(x_0,h(x_0)}$. I concluded that the $V_n$ constitute a neigborhood basis for the graph at $(x_0,h(x_0))$, but this is wrong. As an example take $h(x) = 0$ for $x \le 0$ and $h(x) = \sin(1/x)$. For $x_0 = 0$ you can see that my argument fails. – Paul Frost Mar 18 '21 at 09:08
  • Paul Frost, Thank you very much. – 00GB Mar 18 '21 at 11:21

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To avoid the question of whether a function $h$ satisfying your assumptions exists, let's just consider a dense, connected set $H\in\Bbb R^2$.

If $f_1:\Bbb R\to(0,1)$ is a homeomorphism, then so is $f:(x,y)\mapsto(f_1(x),y)$, so $K=f(H)$ is connected and dense in $(0,1)\times\Bbb R$. The point $(0,0)$ is a limit point of $K$, because every open neighborhood of $(0,0)$ contains an open subset of $(0,1)\times\Bbb R$, which must contain an element of $K$ by density.

Let $R=\{(x,0):x\le0\}$. You're interested in whether $K\cup R$ is connected. Since $K$ and $R$ are each connected, the only potential disconnection to check is $\{K,R\}$. But $R$ is not open because it contains $(0,0)$ and every open set containing $(0,0)$ includes elements of $K$. So $K\cup R$ is indeed connected.

Karl
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  • Kari, Thank you for getting back to the question again. on little thing the $K\cup R$ is not the whole for $g$. Consider, $M={x\colon x\geq o}$. I think I got the same thing, your argument is $K\cup R \cup M $ has the point $(0,0)$ as limit point. but why this will ensure the connectedness. – 00GB Mar 16 '21 at 13:05
  • Kari, I think I got it because .Lemma: If $Y$ is a subspace of $X$, a separation of $Y$ is a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point of the other. The space Y is connected if there exists no separation of Y. – 00GB Mar 16 '21 at 13:42