general question ( open subsets of intervals)

[Meital]

I have a general question on properties of intervals from a topological point of view( intervals in any ordered , linear continuum space, not only reals), so if we have a non-empty open set A in [a,b], and let b be in A for example, and let x < b , x in A. I need to prove that there exists an interval (y,x] in A. I believe that the fact that A is open and the betweenness of A has to do something with that proof. Again, A is any subset of [a,b], not necessary a convex subset. Will someone help me with this, please? Thanks in advance.

[Meital]

A is a non-empty subset of [a,b], where [a,b] is a subset of an ordered, linear continuum space. Now that A is non-empty, let's assume b and x are 2 elements of A.

Now knowing that x is in A, and that A is open, can we say that there exists y < x such that (y,x] is in the subset A? If so, how to show that.

 

This claim is part of a proof I am doing to prove that [a,b] is connected. So I assumed that 2 non-empty disjoint open sets of [a,b], call them A and B, cover [a,b], ( as you see I am doing proof by contradiction), and the claim stated above ( my question) is where I am stuck at..so can you help me please?

[Meital]

Basically it is the same Q I have on page 153 of James R.Munkres book, Topology 2nd edition. My question is on Theorem 24.1 in case 1, how does it follow that there exists some interval of the form (d,c] in B_o, just from knowing that B_o is open in [a,b]?

[Dr. Casey]

Hmm... alright. Take A and put it on top of B. Take x and square it. After squaring x, multiply A times B. The result (AB) goes into the d position, and x goes into the c position. Now repeat the process using the first problem, only insert the denominator resulting from the second question into the first number of the first problem. Now stick AB from the first problem into a, and x from the first problem into b.