- #1
dark904
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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.