# What goes wrong when students interchange “there exists” and “for all” randomly? How to fix this?

I think, it is a very common problem that some students have huge problems with definitions when there appears a quantification.

Some examples:

• Of course the sequence is bounded, because every part of the sequence is a real number and this real number is bounded by itself.
• Look, here is a finite covering for this set, hence it is compact.
• A lot more examples related to $\delta$-$\epsilon$ criterion, etc.
• Also, many students don't quantify new introduced variables which leads then to bigger problems during their proofs.

What goes wrong if students interchange (or leave open) the quantification in the definitions? How can this be fixed or at least improved?

• You're asking about all student confusion related to interchanging quantifiers? There is a lot of work in this area; plenty is easily found through google scholar. What specific scenario are you concerned about and what have you already found in the literature? – Benjamin Dickman Apr 8 '14 at 9:32
• You should never explain this with "every student has a mother" sort of analogy. These analogies will only confuse people further. We had an exam question to explain the difference between two such statements without a rigorous proof. The answers were almost all half-assed explanations "Uhh... every pot has a lid, but there is no one lid for all the pots so.. uhh.."; stick to definitions and explain to students that they should also verify their intuition with a step-by-step analysis from the definitions. – user508 Apr 8 '14 at 17:27

One way to increase awareness of this issue is to explore what happens if the quantors are changed from exists to for all, and vice versa.

On math.SE, there was a question which dealt precisely with this

The following exercise is supposedly from Makarov's Selected Problems in Real Analysis:

Describe the set of functions $f: \mathbb R \rightarrow \mathbb R$ having the following properties ($\epsilon, \delta,x_1,x_2 \in \mathbb R$) :

a) $\forall \epsilon ,\qquad\qquad \exists \delta>0 ,\qquad |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

b) $\forall \epsilon >0 ,\qquad \exists \delta ,\qquad \qquad |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

c) $\forall \epsilon >0 ,\qquad \exists \delta>0 ,\qquad (x_1-x_2) < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

d) $\forall \epsilon >0 ,\qquad \forall \delta>0 ,\qquad |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

e) $\forall \epsilon >0 ,\qquad \exists \delta>0 ,\qquad |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|>\epsilon$

f) $\forall \epsilon >0 ,\qquad \exists \delta>0 ,\qquad |x_1-x_2| < \epsilon \Rightarrow |f(x_1)-f(x_2)|<\delta$

g) $\forall \epsilon >0 ,\qquad \exists \delta>0 ,\qquad |f(x_1)-f(x_2)| > \epsilon \Rightarrow |x_1-x_2|> \delta$

h) $\exists \epsilon >0 ,\qquad \forall \delta>0 ,\qquad |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

i) $\forall \epsilon >0 ,\qquad \exists \delta>0 ,\qquad x_1-x_2 < \delta \Rightarrow f(x_1)-f(x_2)<\epsilon$

• Nice exercise! +1 – Mark Fantini Apr 9 '14 at 3:56
• The omission of the quantifiers for $x_1$ and $x_2$ is rather unfortunate. – Gilles Apr 16 '14 at 11:48
• @Gilles: Indeed, it is. However, I kept it this way since it was originally posted like this. – Roland Apr 16 '14 at 12:13