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I'm trying to solve the following exercise:

In each case below, say whether the given statement is true for the universe (0, 1) = {x ∈ R | 0 < x < 1}, and say whether it is true for the universe [0, 1] = {x ∈ R | 0 ≤ x ≤ 1}. For each of the four cases, you should therefore give two true-or-false answers.

  • a. ∀x(∃y(x > y))
  • b. ∀x(∃y(x ≥ y))
  • c. ∃y(∀x(x > y))
  • d. ∃y(∀x(x ≥ y))

From here I can see that a(every number in the universe has a number less than itself) is true in the open interval, but not in the closed. I just can't get my head around why that is. I know that in the open interval 0 is not included, so how can the lowest possible value in that interval(exclusive 0) still have a y that is lower than it? If any one can give me a good explanation for this I would really appreciate it.

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J.s
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1 Answers1

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The reason is following. I claim that in $(0,1)$, for any $x\in(0,1)$, there exists $y\in(0,1)$ such that $y<x$. One way to prove my claim is to construct $y$ for any $x$, and I can do so by taking $y=x/2$.

I cannot do the construction (and hence prove similar claim) in $[0,1]$, since for $x=0$, $y\in[0,1]$ such that $y<x$ does not exist.

As a side remark, your 'lowest possible value in that interval(exclusive 0) still have a y that is lower than it' is not what the property at hand is. The property at hand says that given $X$, for any $x\in X$, there exists $y\in X$ such that $y<x$.

Jan
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