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If i suppose that $I$ is an interval but it is not connected,i.e there existe two open sets on $I$: $U,V$ such that $I=U\cup V$ et $U\cap V=\emptyset$

Let $x\in U$ and $y\in V$ (for example $x<y$) and let $B_x=\{z\in U, [x,z]\subset U\}$

Then $B_x$ is bounded by y .

My question is why $B_x$ is bounded by $y$ ?

Thank you

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You're wrong in your assumptions. Saying that $I$ is not connected means there exist two open sets $U$ and $V$ such that

  • $I\subseteq U\cup V$
  • $(U\cap I)\cap(V\cap I)=\emptyset$
  • $U\cap I\ne\emptyset$ and $V\cap I\ne\emptyset$

Or you have to say that $U$ and $V$ are open sets in $I$ (with the relative topology).

Fix $x\in U\cap I$ and $y\in V\cap I$. It's not restrictive to assume $x<y$ (otherwise exchange the roles of $U$ and $V$).

Then $B_x=\{z\in I:[x,z]\subseteq U\}$ is bounded by $y$, because if $z\in B_x$ and $z>y$, then $[x,z]\subseteq U$ implies $y\in [x,z]$ and so $y\in U$: this is a contradiction.

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  • $\begingroup$ Thank you, after that theu say let $M=\sup B_x$ then $x\leq H\leq y$, As $I$ is an interval $M\in I$ moreover $B_x\subset U\subset I \Rightarrow M\in \overline{B_x}\subset \overline{U}$ but $U=C_{I}V$ closed $\Rightarrow M\in U$ As $U$ is open $\exists \varepsilon>0, ]M-\varepsilon,M+\varepsilon[\subset U$ the conclusion follows from the definition of sup. I don't understand this part of proof $\endgroup$ – Vrouvrou Mar 18 '15 at 13:37
  • $\begingroup$ Please why $U=C_{I}V$ ? thank you $\endgroup$ – Vrouvrou Jan 13 '16 at 15:55
  • $\begingroup$ @Vrouvrou What's $C_IV$? And what does this have to do with the proof I presented? $\endgroup$ – egreg Jan 13 '16 at 16:03
  • $\begingroup$ ah sorry , there is no relation with your answer , i just want to know why U and V are open and closed in I sorry $\endgroup$ – Vrouvrou Jan 13 '16 at 16:21
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    $\begingroup$ @Vrouvrou Please! $(M-\varepsilon,M+\varepsilon)\subseteq U$; on the other hand, by definition of supremum, there exists $z\in B_x$ such that $M-\varepsilon<z\le M$. So $[x,z]\subseteq U$; therefore $[x,z]\cup(M-\varepsilon,M+\varepsilon/2]\subseteq U$. What's that union? Of course it contains $[x,M+\varepsilon/2]$. $\endgroup$ – egreg Jan 14 '16 at 22:01
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Because if $z\geq y$, then $y\in [x,z]$, this means that $[x,z]$ cannot be a subset of $U$

(because it contains at least one element, $y$, from $V$, and since $y\in V$, we know that $y\notin U$).


For part two, you show that there exists such an $\epsilon$ that $(M-\epsilon, M+\epsilon)$ is a subset of $U$. This means that $M$ is not the upper bound of $B_x$, because $[x,M+\epsilon)$ is a subset of $U$, so, for example, $M+\frac{\epsilon}{2}$ is in $B_x$ (because $[x,M+\frac\epsilon2] \subset [x, M+\epsilon)\subset U$).

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  • $\begingroup$ Thank you, after that theu say let $M=\sup B_x$ then $x\leq H\leq y$, As $I$ is an interval $M\in I$ moreover $B_x\subset U\subset I \Rightarrow M\in \overline{B_x}\subset \overline{U}$ but $U=C_{I}V$ closed $\Rightarrow M\in U$ As $U$ is open $\exists \varepsilon>0, ]M-\varepsilon,M+\varepsilon[\subset U$ the conclusion follows from the definition of sup. I don't understand this part of proof $\endgroup$ – Vrouvrou Mar 18 '15 at 13:48
  • $\begingroup$ @Vrouvrou What's $H$ in $x\leq H\leq y$? Also, rather than in a comment, post what you just posted as an edit to your question, making it clear that that's part two of your question. $\endgroup$ – 5xum Mar 18 '15 at 13:51
  • $\begingroup$ @Vrouvrou See the edit to my answer for part two. $\endgroup$ – 5xum Mar 18 '15 at 13:54
  • $\begingroup$ it is not $H$ sorry it is $M$, can you explain me the part "moreover $B_x\subset U\subset I \Rightarrow M\in \overline{B_x}\subset \overline{U}$..." $\endgroup$ – Vrouvrou Mar 18 '15 at 14:03
  • $\begingroup$ please why the fact that $B_x\subset U\subset I$ impliese that $M\in \overline{B_x}\subset \overline{U}$ ? $\endgroup$ – Vrouvrou Mar 18 '15 at 17:11
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Suppose $B_{x}$ were not bounded by $y$, i.e., there is some $z$ such that $y \leq z$ and $[x,z] \subset U$. Then since $y \in [x,z]$, we have $y \in U$, which contradicts that $y \in V$ and $U \cap V = \emptyset$.

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  • $\begingroup$ Thank you, after that theu say let $M=\sup B_x$ then $x\leq H\leq y$, As $I$ is an interval $M\in I$ moreover $B_x\subset U\subset I \Rightarrow M\in \overline{B_x}\subset \overline{U}$ but $U=C_{I}V$ closed $\Rightarrow M\in U$ As $U$ is open $\exists \varepsilon>0, ]M-\varepsilon,M+\varepsilon[\subset U$ the conclusion follows from the definition of sup. I don't understand this part of proof $\endgroup$ – Vrouvrou Mar 18 '15 at 13:48

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