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.
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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