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 Oct 6 '16 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 Oct 6 '16 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 $|f(x)-L|<\epsilon$ $\endgroup$ – Callus Oct 6 '16 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. $\endgroup$ – Andres Mejia Oct 6 '16 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) $ $\endgroup$ – Andres Mejia Oct 6 '16 at 11:15

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.