We know that surjectivity is defined as follows:

Definition (Surjective Function): Let $f:A\subset\mathbb{R}\rightarrow B\subset\mathbb{R}$. $\forall y\in B$, $\exists x\in A$ such that $y=f(x)$, then $f$ is called a surjective function.

My inquiry rise while defining the limit:

Definition (Limit): Let $f:A\subset\mathbb{R}\rightarrow B\subset\mathbb{R}$. The following statements are equivalent.

  • $\forall \epsilon>0$, $\exists\delta=\delta(\epsilon)>0$ whenever $0<|x-x_0|<\delta$ then $|f(x)-L|<\epsilon$.
  • $\lim_{x\rightarrow x_0} f(x)=L$.

My question is: Is the relation $\delta=\delta(\epsilon)$ surjective? If not, is there any condition which makes it surjective?

  • 1
    $\begingroup$ Why do you believe that $\delta$ is a function of $\epsilon$? Please remember that being a function of has a technical meaning in mathematics, while in language it may be interpreted in a more relaxed way. $\endgroup$
    – Siminore
    Commented Oct 6, 2016 at 11:09
  • $\begingroup$ Your definition of surjective is wrong. Think about it like a picture. You have a blob of points for your domain, and a blob of points for your codomain, and the map $f$ assigns each point in the domain to a single point in the codomain. Surjective means every point in the codomain has something from the domain being mapped to it. That's different than what you said. $\endgroup$
    – layman
    Commented Oct 6, 2016 at 11:11
  • $\begingroup$ Your definition of surjective is backwards. It should say for every element of the codomain ( not range ) there is an element of the domain that maps to it. As @Siminore points out, the definition of limit tells you that for every $\epsilon$ there is at least one $\delta$ satisfying that property, so there is not necessarily one choice of $\delta$ for each $\epsilon$, so it's not necessarily a function. Lastly, your definition of limit is incorrect. It should say if $|x-x_0|<\delta$ then $|f(x)-L|<\epsilon$ $\endgroup$ Commented Oct 6, 2016 at 11:13
  • $\begingroup$ Your definition for subjectivity is incorrect. The range is precisely the set of points in the domain so that the function is surjective on it. $\endgroup$ Commented Oct 6, 2016 at 11:13
  • 1
    $\begingroup$ Perhaps the following characterization of continuity will clarify: For all $B (f (x),\epsilon) $ there exists some neighborhood of $x$ so that $f (B (x,\delta) \subseteq B (f (x),\epsilon) $ $\endgroup$ Commented Oct 6, 2016 at 11:15

1 Answer 1


Well, your question is a bit unclear, and as often is the case with such questions, the answer is a resound "Maybe!"

Interpretation one: functions.

The problem is that $\delta$ isn't a function of $\epsilon$. Can we fix this?

The answer is yes: look for the biggest possible $\delta$. Namely, if $f$ is continuous and $x_0$ is a point in the domain of $f$, we define $\delta_{x_0}(\epsilon)$ as $$\sup\{\delta: \forall x(0<\vert x-x_0\vert<\delta\implies \vert f(x)-L\vert<\epsilon)\}$$ (where $L=\lim_{x\rightarrow x_0}f(x)$).

Exercise: show that $\delta_{x_0}(\epsilon)$ actually "works" for $\epsilon$ - that is, $$\forall x(0<\vert x-x_0\vert<\delta_{x_0}(\epsilon)\implies \vert f(x)-L\vert<\epsilon).$$

Now for each $x_0$, $\delta_{x_0}$ is a map from $\mathbb{R}_{>0}\cup\{\infty\}$ to $\mathbb{R}_{>0}\cup\{\infty\}$ (exercise: think about why it's convenient to have "$\infty$" here), and we can ask if this map is surjective. Unfortunately, the answer is no: e.g. if $f(x)=0$ for all $x$, then $\delta_{x_0}(\epsilon)=\infty$ for all $x_0$ and all $\epsilon$.

Interpretation two: relations.

Another approach is to talk about surjective total relations. A total relation $R$ between $A$ and $B$ is basically a multivalued function: for each $a\in A$, there is at least one $b\in B$ with $R(a, b)$. There is a natural "delta-relation" given by $\Delta_{x_0}(a, b)$ iff $$\forall x(0<\vert x-x_0\vert<b\implies \vert f(x)-L\vert<a)$$ (in words, $\delta=b$ "works for" $\epsilon=a$). If $f$ is defined on all of $\mathbb{R}$, then it is then true that "every $\delta$ works for some $\epsilon$", and this is a good exercise (HINT: extreme value theorem).

Note: here I'm restricting to finite $\epsilon$ and $\delta$. It's a good exercise to think about how the previous claim needs to change depending how infinite values of $\epsilon$ and $\delta$ are allowed in the picture.

Note, in both interpretations, the dependence on the point $x_0$. This can't be done away with; in general, the "$\epsilon-\delta$ structure" of a function at one point is very different than at a different point. But see here for more details.


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