Continuity at $+\infty$ for the function defined by $f(0)=\infty$ and $f(x)=1/x$ for $x \in (0,10]$.

Question

Let the domain of the function $f(x)$ be $[0, 10]$, and its range be the extended real numbers (including +$\infty$ and -$\infty$). Define:

$ f(x) = \left\{ \begin{array}{lr} 1/x & : 0 < x \leq 10\\ +\infty & : x = 0 \end{array} \right. $

Question: why is $f(x)$ not continuous at $0$?

Context: there is a theorem that states a function is uniformly continuous on $(a,b)$ iff it can be extended to a continuous function on $[a,b]$. Because $1/x$ is not uniformly continuous on $(0,10)$, the extension function $f(x)$ defined above must not be continuous.

Answer 1

All you need is the $\epsilon-\delta$ definition of continuity. If $x=0$ and you want $\epsilon = 1$ what will $\delta$ be? $\epsilon-\delta$ definition is applicable to any metric space, not only the real line.

The confusion is arising from $lim_{x \to 0} f(x) = \infty$. This is a statement very different from $lim_{x \to 1} f(x) = 1$, for example. In the former case there is no converging going on. The values of $f(x)$ never get particularly close to $\infty$. This is a notation only used for real numbers. Using the topological definition of a limit we would say that $lim_{x \to 0} f(x)$ does not exist.

Answer 2

Suppose $f$ was continuous at zero. Then there exists some $\delta>0$ s.t. $0<\left|x-0\right|=\left|x\right|<\delta$ implies $\left|f\left(x\right)-\infty\right|<1$. Spot the contradiction?

Answer 3

There is a sense in which this is continuous, using the "obvious" topology that one would get with adding $\infty$, in particular $(r,\infty]$ would be open (and you consider the topology generated by the standard topology plus these new "balls" around $\infty$). So $f(0)=\infty$, and we may consider an open ball at $\infty$, say $(r,\infty]$. Well $f^{-1}(r,\infty] = [0,1/r) \cap [0,10]$ which is open.