Basic Real Analysis: Are Open Intervals Always Roomy?

[dark904]

Homework Statement

This problem starts with a definition,
A set S [itex]\subseteq[/itex] R is said to be roomy if for every x [itex]\in[/itex] S, there is a positive distance y > 0 such that the open interval (x - y, x + y) is also contained in S.

Problems based on this definition:
a) Let a < b. Prove that the open interval (a , b) is roomy.
b) Suppose that A and B are roomy sets. Prove or disprove: A [itex]\cup[/itex] B is a roomy set.

2. The attempt at a solution

There are a couple of immediate consequences of this definition that I can think of. The most important ones are that closed intervals cannot be roomy and that the empty set is roomy (since this would make the first part of the implication false).

a) This part seems obvious but I can't figure out how to lay this out as an actual rigorous proof. It's apparent to me that since we are in R and (a,b) is open, there will always be some number between the ends of the interval and x. How do I say this concisely and does it require a subproof?

b) I can't think of a counterexample so I'm inclined to say that this is true.

Here is my attempt at a proof:
Since A and B are both roomy sets, they must both be open intervals, as a closed interval would contradict the definition of roominess. Since A and B are open intervals, A [itex]\cup[/itex] B is either the empty set or also an open interval. If A [itex]\cup[/itex] B is the empty set, then it is roomy and if it is an open interval then it is also roomy, therefore A [itex]\cup[/itex] B is roomy.

 

[Dick]

Homework Statement

This problem starts with a definition,
A set S [itex]\subseteq[/itex] R is said to be roomy if for every x [itex]\in[/itex] S, there is a positive distance y > 0 such that the open interval (x - y, x + y) is also contained in S.

Problems based on this definition:
a) Let a < b. Prove that the open interval (a , b) is roomy.
b) Suppose that A and B are roomy sets. Prove or disprove: A [itex]\cup[/itex] B is a roomy set.

2. The attempt at a solution

There are a couple of immediate consequences of this definition that I can think of. The most important ones are that closed intervals cannot be roomy and that the empty set is roomy (since this would make the first part of the implication false).

a) This part seems obvious but I can't figure out how to lay this out as an actual rigorous proof. It's apparent to me that since we are in R and (a,b) is open, there will always be some number between the ends of the interval and x. How do I say this concisely and does it require a subproof?

b) I can't think of a counterexample so I'm inclined to say that this is true.

Here is my attempt at a proof:
Since A and B are both roomy sets, they must both be open intervals, as a closed interval would contradict the definition of roominess. Since A and B are open intervals, A [itex]\cup[/itex] B is either the empty set or also an open interval. If A [itex]\cup[/itex] B is the empty set, then it is roomy and if it is an open interval then it is also roomy, therefore A [itex]\cup[/itex] B is roomy.

For a) pick an x in (a,b) and tell how to define the corresponding y. That would be a proof. For b) you seem to think the only subsets of R are open and closed intervals. Give some examples that aren't and then rethink the proof. The union of two intervals isn't necessarily an open interval!

 

[dark904]

Hi Dick, thanks for the speedy response.

For part b, I understand that there are subsets of R that aren't intervals, however by the problem's definition of "roominess," the only subsets of R which are roomy are open intervals, as it is impossible for a y > 0 such that (x -y , x + y) [itex]\subseteq[/itex] S to exist in a set with a closed end or in a singleton set or in a set consisting of a finite collection of reals, so the only sets relevant to the proof are open intervals because the problem states that A and B are roomy.

 

[Dick]

Hi Dick, thanks for the speedy response.

For part b, I understand that there are subsets of R that aren't intervals, however by the problem's definition of "roominess," the only subsets of R which are roomy are open intervals, as it is impossible for a y > 0 such that (x -y , x + y) [itex]\subseteq[/itex] S to exist in a set with a closed end or in a singleton set or in a set consisting of a finite collection of reals, so the only sets relevant to the proof are open intervals because the problem states that A and B are roomy.

Hi dark904. The set (0,1)U(2,3) is roomy. It's NOT an open interval. Don't try to claim that in your proof. Use the definition of 'roomy' in your proof. Don't try to claim everything is some kind of interval.