The exercise is as follows: Given a function $f : D \mapsto \mathbb R$ Are the following conditions stronger, weaker or not comparable with continuity?

$$\forall a \in D, \exists \epsilon \gt 0, \exists \delta \gt 0, \forall x \in D: |x-a| \lt \delta \Longrightarrow |f(x) - f(a)|\lt \epsilon $$

The solution that was given is:

It is weaker because $\forall \epsilon \gt 0 P(\epsilon) \Rightarrow \exists \epsilon \gt 0 : P(\epsilon)$

As an example $f(x)= \begin{cases} 0 & x \ne 0\\ 1 & x = 0 \end{cases}$

Now (c), because $\epsilon = 2, a\in D, \delta = 1, \forall x \in D : |x-a| \lt 1 \Longrightarrow |f(x) -f(a)|\le 1 \lt 2 $

What I understand from this "solution" is that because of the quantifier $\exists$ is weaker than the quantifier $\forall$ the statement is weaker. Ok, from here on I don't understand it.

The example is a discontinuous function, ok. But I don't see the connection between the statement and the example. At least I'm not 100 % sure of it because I don't see why the statement suggests anything discontinuous.

  • $\begingroup$ "Quantifier", not "quantor". $\endgroup$ Commented Jan 16, 2018 at 18:37
  • $\begingroup$ Ok, in my language it is quantor. My apologies $\endgroup$ Commented Jan 16, 2018 at 18:39
  • $\begingroup$ Why did you vote to close my question if I may ask? $\endgroup$ Commented Jan 16, 2018 at 18:40
  • $\begingroup$ I'll edit it, sir. $\endgroup$ Commented Jan 16, 2018 at 18:59
  • $\begingroup$ I don't know who voted or why; I think this is a well written question. $\endgroup$
    – StackTD
    Commented Jan 16, 2018 at 19:07

2 Answers 2


The function in the example satisfies the first statement since there exists an $\epsilon$ which makes the statement true; they show this by giving explicit values that indeed make it work.

However, that function is clearly discontinuous. The key is that the given statement is indeed weaker than the one defining continuity since, in that definition, it doesn't suffice to find one specific value of $\epsilon$ for which the rest of the statement holds, it needs to hold for all possible (positive) values of $\epsilon$. For the function in the example, it fails for any $\epsilon$ smaller than 1.

Letting go of this (mathematical) context, it's useful to see how for any statement "there exists (...)" is indeed weaker than "for all (...)"; which they wrote symbolically:

It is weaker because $\forall \epsilon \gt 0 P(\epsilon) \Rightarrow \exists \epsilon \gt 0 : P(\epsilon)$

If you're in a class with at least one student, take the property "is male" and consider the statements:

  1. there exists a student $S$ in class: "$S$ is male"
  2. for all students $S$ in class: "$S$ is male"

It should be clear that if the second statement is true, then the first is as well; but necessarily the other way around (e.g. a class with at least one man and at least one woman).


The statement is "stronger" in the following sense: A continuous function will always satisfy that statement, but it is not necessarily the case that a function that satisfies that statement will be continuous.

The example is intended to demonstrate this point since it satisfies the statement but is not continuous.


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