Real analysis proof

**Danneedshelp** · November 17th 2009, 07:47 PM

Prove: a) Let $\rho$ be a transitive relation on the interval $[a,b]$ . If each $x\in[a,b]$ has a neighborhood $N_{x}$ such that $u\rho\\v$ whenever $u\in[a,x]\cap\\N_{x}$ and $v\in[x,b]\cap\\N_{x}$ , then $a\rho\\b$ .

b) Use the above result to prove Cantors Nested Intervals Property (the intersection of a family of closed sets is non-empty).

A: Here is kind of an outline to part (a):

Let $x\in[a,b]$ be arbitrary. Since $[a,b]$ is compact we know $b=max([a,b])=sup([a,b])$ . Now, by our hypothesis, we can find a neightborhood of $x$ such that if $u\in[a,x]\cap\\N_{x}$ and $v\in[x,b]\cap\\N_{x}$ , then $u\rho\\v$ . So, there are two cases; first, $N_{x}$ is an interior point of the interval; second, $N_{x}$ is not an interior point, so a portion of the neighborhood may be outside the closed interval ( $x$ could even equal $b$ ). In either case, we can find a transitive relation [tex]\rho[/mathp such that $u\rho\\v$ implies $a\rho\\b$ , since we know [tex]b=sup([a,b]).

I am having troubles formalizing the proof and organizing my thoughts as you may be able tell.

b) I am stuck on this one even though my teacher offered the hint, "consider the relation $u\rho\\v\$ iff $u\leq\\v$ and $(\exists\\n\in{\mathbb{N}}$ s.t. $[u,v]\cap[a,b]=\emptyset$ .

Thanks

**Danneedshelp** · November 18th 2009, 07:38 AM

Any ideas on the second part? I am stuck.

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