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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

Nehushtan
2013-04-06 00:53:17

The statements are not the same. In statement 1, one single d holds for all a; in statement 2, different a may be associated with different d. Statement 2 only says that there exists at least one d for each a.


This distinction is the difference between continuity and uniform continuity. Let I be a real interval and a real-valued function on I. Then:

(i)  is continuous on I iff
(∀ε>0)(∀xI)(∃δ>0)(∀yI)(|xy|<δ⇒|(x)−(y)|<ε)

(ii) is uniformly continuous on I iff
(∀ε>0)(∃δ>0)(∀xI)(∀yI)(|xy|<δ⇒|(x)−(y)|<ε)

The two definitions are different. In (i) δ depends on x; different δ may need to be chosen for different x. In (ii), one single δ has to work for all x.

stapel
2013-04-05 23:51:23

FlashBurn wrote:

Statement 2: For all a ∈ A there exists d ∈ D such that H(a, d).

This one I can't figure out. Does it mean that for every a there exists a unique d...or does it mean that d's can be shared by some a's...?

Since Statement 2 doesn't say "there exists a unique d", I would interpret this in the same manner as for Statement 1; namely, that there exists some element d for each a. The element d doesn't have to be unique (a different d for each a). wink

FlashBurn
2013-04-05 14:21:35

I'm trying to understand what happens when quantifiers in the prediate are swapped.

Let's assume the following:
Two sets A and a D and a predicate H(a, d), where a ∈ A and d ∈ D.

Statement 1.

There exists d ∈ D for all a ∈ A such that H(a, d).

This one I can figure out. It means that there exists a single d for all a such that H(a, d).

Statement 2.

For all a ∈ A there exists d ∈ D such that H(a, d).

This one I can't figure out. Does it mean that for every a there exists a unique d, i.e a1 - d1, a2 - d2, a3 - d3 etc. or does it mean that d's can be shared by some a's, i.e. a1 - d1, a2 - d2, a3 - d1, a4 - d2  etc.

Any help is appreciated.

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