Can $f(x)$ defined on an open interval have limits at the endpoints? Can $f(z)$ defined on an open disk have limits at the boundaries?

Question

• For suppose $f$ a function from $(0,1)$ onto $(0,1)$ defined by $f(x)=x$, then is the limit of $f$ exist as $x$ tends to $0$ or $1$, I know there only one side limit is finitely defined, but from the other side, it is not defined. Now my question is that is the limit exist still (as vacuously)?

• Also for a linear complex function $f$ defined on a region $\{ z : |z|<1\}$, does $f$ have limit value at the boundary points? Since, we can't reach the boundary points from all the directions, as it is not defined! Now what about the existence of limit, is it vacuously exist from all the othe directions(not defined in domain!).

And my last question is little different from above.

• For suppose limit of function have existed at all points of its domain. Then is it possible to find such function that have limit value as limit point of codomain or outside of codomain?
  Please anyone help , any kind of help would be appreciated!

Answer 1

I think your question is related to an extension of a continuous function to limit points of its domain. This problem is investigated, for different topological spaces, see the references. A general problem is the following. Given a continuous function $f$ from a subspace $D$ of a topological space $X$ to a topological space $Y$, whether $f$ can be extended to a continuous function from $X$ to $Y$.

From now we shall consider a particular interesting case when $D$ is dense in $X$. In this case each point of $X\setminus D$ is a limit point of $X$. If $Y$ is Hausdorff, then is well-known (see, for instance [Eng, Theorem 1.5.4]) and easy to show that if an extension $\bar f$ of $f$ exists then it is unique. It follows that $f’(X)\subset f(X)$ for any extension $f’$ of $f$ iff $\bar f(X)\subset f(X)$ for some extension $\bar f$ of $f$.

A characterization of extendable maps $f$ is provided in lemmas in this my answer. They follows that when both spaces $X$ and $Y$ are metric (for instance, when they are subspaces of the real line or the plane), $f$ can be extended from $D$ to $X$ is iff for each sequence $\{x_n\}$ of points of the set $D$ convergent in $X$, a sequence $\{f(x_n)\}$ is also convergent in $Y$. If $Y$ is a complete metric space then the latter condition is equivalent to $\{f(x_n)\}$ is a Cauchy sequence. In particular, if $f$ is uniformly continuous then it is extendable. The completeness of $Y$ is essential here. Indeed, if $X=[0,1]$ and $D=Y=(0,1)$ then although the identity function $f$ from $D$ to $Y$ keeps Cauchy sequences but it cannot be extended to $X$.

An other important case takes place when $D$ is Tychonoff, $Y=[0,1]$ and each continuous map from $D$ to $Y$ can be extended to $X$. Then $X$ is equivalent to a Stone-Čech compactification of $D$, see [Eng, Corollary 3.6.3]. This follows that each continuous map from $D$ to any compact space $Y$ can be extended to $X$.

References

[Ano] Anonymous et al., An extension of a continuous function onto the closure.

[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

[Pau] Paul et al., Extending continuous function $D\to I$, where $D$ is a dense subspace of a separable Tychonoff space.