Question

What is the union of all of the closed intervals contained in the open interval (0, 1)? justify your answer.

Answers

  1. After then, there must be some closed interval A between 0 and 1 such that x is less than A. As a result, x must fall between 0 and 1, as was intended.
    This is further explained below.

    What is the union of all of the closed intervals contained in the open interval (0, 1)?

    Generally, Let A represents the set of all of the closed intervals that are included in (0, 1). Our contention is that A = (0, 1). To demonstrate this, let’s begin by demonstrating that the left-hand side contains the right-hand side in its entirety.
    Let x ∈ (0, 1). Our goal is to demonstrate that there is at least one A A such that x A. We may define A to be the closed interval [x, x] that contains the single point x. Alternatively, if you don’t wish to state that a point is an interval, we can take A to be something like [x/2, x] or [x/2, (x + 1)/2] as an alternative.
    In conclusion, because x > A > A, we may deduce that x > A A A. Now we will perform the confinement in the other direction. Let’s say that x > AA A.
    If this is the case, there must be some closed interval A between 0 and 1 such that x falls inside A. Therefore, x should be between 0 and 1, as required.
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