Let $f$ be a continuous function defined on $\mathbb{R}^n$. The range of $f$ is in the extended real numbers. Is $f<\infty \ \forall x\in \mathbb{R}^n$ ? And why?

  • $\begingroup$ Is this really the question you mean to ask? If so, the answer is clearly yes: the range of the function is the set of real numbers, and $\infty$ is not a real number. All real numbers are finite. The function might be unbounded - so that it takes on values as large as you might imagine. $\endgroup$ Mar 24, 2016 at 13:49
  • $\begingroup$ What if the range is in extended real numbers, as we are studying measure theory. $\endgroup$
    – Allitee
    Mar 24, 2016 at 13:51
  • $\begingroup$ Then what about the constant function $f:x\mapsto\infty$? $\endgroup$
    – MPW
    Mar 24, 2016 at 13:55
  • $\begingroup$ Do you have a definition of what it means to be continuous at a point $x_0$ such that $f(x_0) = \infty$? $\endgroup$
    – angryavian
    Mar 24, 2016 at 13:56
  • $\begingroup$ Then you need to specify the topology on the extended real numbers. In any case the function that has constant value $\infty$ will be continuous. $\endgroup$ Mar 24, 2016 at 13:56

1 Answer 1


In the "usual" topology on the extended real numbers $[-\infty, \infty]$, a neighborhood of $\infty$ is a set containing some interval $(a, \infty] = (a, \infty) \cup \{\infty\}$ with $a$ real, and similarly for neighborhoods of $-\infty$.

If that's true in your setting, $[-\infty, \infty]$ is homeomorphic to a closed, bounded interval of real numbers. For instance, the hyperbolic tangent function $\tanh$ is a homeomorphism from $[-\infty, \infty]$ to $[-1, 1]$. Consequently, asking whether a continuous function can achieve the value $\infty$ is no more mysterious than asking whether a continuous, real-valued function can achieve an absolute maximum.

As MPW and Ethan Bolker note, the constant function with value $\infty$ is continuous. Non-constant continuous functions can achieve the values $\infty$ and/or $-\infty$, as well. For example, if $n = 1$, then

  • $f(x) = \frac{1}{x^{2}}$ (extended by $f(0) = \infty$) is continuous, since $\lim\limits_{x \to 0} \frac{1}{x^{2}} = \infty$.

  • $f(x) = \frac{1}{x}$ (extended by $f(0) = \infty$) is not continuous, since $\lim\limits_{x \to 0^{\pm}} \frac{1}{x} = \pm\infty$ (i.e., the one-sided limits exist as extended real numbers, but are not equal).

And so forth.

  • $\begingroup$ Hi, I am wondering how can I show $f(x)=\frac 1 {x^2}$ is continuous at $x=0$? $\endgroup$
    – Allitee
    Mar 24, 2016 at 14:39
  • $\begingroup$ The customary calculus definition of "$f(x) \to \infty$ as $x \to c$" reads: "For every real number $a$, there exists a $\delta > 0$ such that if $0 < |x - c| < \delta$, then $f(x) > a$." The definition of continuity can then be extended to read "If $f(c) = \infty$, then $f$ is continuous at $c$ if $f(x) \to \infty$ as $x \to c$." $\endgroup$ Mar 24, 2016 at 15:43

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