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If i suppose that $\mathbb{R}$ is not connected, then there exists $A\subset \mathbb{R}$ is closed and open , where $\emptyset\neq A\neq \mathbb{R}$

How to continue with this methode ?

Edit: Let $x\notin A$ we suppose that $A\cap ]-\infty,x]\neq \emptyset$ (or we work with that set $A\cap [x,+\infty[$ )

this set is closed and open sice $x\notin A$ and it is bounded from upper by $x$

so there exists $y=\sup (A\cap ]-\infty,x])$, As $A\cap ]-\infty,x]$ is open; $\exists \varepsilon>0, ]y-\varepsilon, y+\varepsilon[\subset (A\cap ]-\infty, x])$

Where is the contradiction please

Thank you

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Suppose that $x\notin A$, then $A\cap(-\infty,x]$ is closed and $A\cap(-\infty,x)$ is open. But since $x\notin A$, this is the same set.

So we can assume that $x=\sup A$. Now derive a contradiction, since $\sup A\in A$ for all closed $A$.

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  • $\begingroup$ what you mean by :"since $x\notin A$ this is the same set ? $\endgroup$ – Vrouvrou Feb 13 '17 at 18:28
  • $\begingroup$ $A\cap(-\infty,x)=A\cap(-\infty,x]$. $\endgroup$ – Asaf Karagila Feb 13 '17 at 18:29
  • $\begingroup$ ok, but i don't understand the idea we must find contradiction with what ? and what we do with the fact that $A\cap (-\infty,x]$ is closed ? $\endgroup$ – Vrouvrou Feb 13 '17 at 18:32
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HINT: Pick some $x\in\mathbb{R}\setminus A$ and $y\in A$. Suppose WLOG that $x<y$; think about the interval $(x, y)$. You want to find some point where, going from $x$ to $y$, we "switch" from $\mathbb{R}\setminus A$ to $A$ (could such a point be in either $A$ or $\mathbb{R}\setminus A$? use the fact that they're both open . . .); so, do you see a way to apply the least upper bound property here?

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  • $\begingroup$ i don't see , i don't understand $\endgroup$ – Vrouvrou Feb 13 '17 at 18:17
  • $\begingroup$ @Vrouvrou What part don't you understand? $\endgroup$ – Noah Schweber Feb 13 '17 at 18:18
  • $\begingroup$ You want to find some point where, going from x to y, we "switch" from R∖A to A (could such a point be in either A or R∖A? use the fact that they're both open . . .); so, do you see a way to apply the least upper bound property here? $\endgroup$ – Vrouvrou Feb 13 '17 at 18:20
  • $\begingroup$ Please what is the contradiction here ? $\endgroup$ – Vrouvrou Feb 13 '17 at 18:36
  • $\begingroup$ @Vrouvrou It's the part in the parentheses. If there were such a point, it would have to be in either $A$ or $A$'s complement; but each of these sets is open, so some neighborhood around this point would be entirely in one set or entirely in the other. But that means it's not a "switching point". Of course that's informal, but it hopefully gives you enough intuition to start in the right direction . . . $\endgroup$ – Noah Schweber Feb 13 '17 at 18:47
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This is a classical result, and you will find many treatments of it in the literature. A hint is that the crucial difference between the reals and the rationals is that the satisfy the upper bound property. So it is natural to consider the supremum/infimum of the sets forming the disconnection. It is technically a bit simpler to show that a segment is connected.

There are alternatives to the classical arguments, explored in this paper. Briefly, connectedness can be phrased in terms of connecting walks, very similarly to how path connectedness is phrased. With one only needs to show that given ant two points in $\mathbb R$ and any scale, a connecting walk exists, which is not hard to do. If resorting to the compactness of a closed segment is allowed, then that part becomes trivial, since the scale may be assumed to be uniform.

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Let $A$ be open and closed with $x\in A$ .

(i). Let $S^+=\{y: y>x\land [x,y)\subset A\}.$ Then $S^+$ is not empty (because $A$ is open, so $A\supset (-r+x,r+x)$ for some $r>0,$ so $r+x\in S^+$).

Suppose $y_+=\sup S^+$ exists in $\mathbb R.$ Then $y_+\in A$

This is because, for any $z\in (x,y_+)$ there is some $y\in (z,s_+]\cap S,$ so $z\in (x,y)\subset A.$.. So $(x,y_+)\subset A,$ and $A$ is closed, so $y_+\in A.$

But now we have a contradiction because $(-r+y_+,r+y_+)\subset A$ for some $r>0,$ so $A\supset [x,y_+]\cup (y_+, r+y_+)=[x,r+y_+),$ implying $\sup S^+\geq r+y_+>y_+=\sup S^+.$

We conclude that $\sup S^+$ does not exist, so $A\supset [x,\infty).$

(ii). Let $(-A)=\{-z: z\in A\}.$ Then $(-A)$ is open and closed. Applying the above argument to $(-A)$ we have $(-A)\supset [-x,\infty)$. Therefore $A\supset (-\infty, x].$

(iii). From (i) and (ii) we have $A=\mathbb R.$

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  • $\begingroup$ why $r+y_+\leq sup S^+ $ ? $\endgroup$ – Vrouvrou Feb 14 '17 at 5:28
  • $\begingroup$ Because $r+y_+>x$ and $A\supset [x, r+y_+)$ .So by the def'n of $S^+$ we have $ r+y_+\in S^+$, which implies $\sup S^+\geq r+y_+.$ $\endgroup$ – DanielWainfleet Feb 14 '17 at 7:55

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