0
$\begingroup$

Essentially, I need to prove that, given a point $x_0$ in $S$ where $S \subseteq \mathbb{R}$, as the $x$ value converges to $x_0$, $f(x)$ converges to only one $f(x_0)$. This is poking at the idea that for every input, $x$, there cannot be more than one output, $f(x)$.

This is something that most of us knew from algebra 1, but I need to prove this statement using the definition of cluster points, continuity, etc.

Cluster point: if $x_0$ is a cluster point, then $\forall \epsilon \gt 0$, $(x_0- \epsilon , x_0+ \epsilon ) \cap (S \setminus {x_0}) \neq \phi $

definition of continuity: $\forall \epsilon \gt 0$, $\exists \delta \gt 0$ such that $0 \lt |x-x_0| \lt \delta$, $x \in S$, implies $|f(x)-f(x_0)| \lt \epsilon$.

I'm not too sure how I would approach this!

$\endgroup$
  • $\begingroup$ Since in Hausdorff spaces limits are unique, there is nothing more to consider. Is f supposed to be continuous? $\endgroup$ – William Elliot Nov 26 '18 at 1:35
0
$\begingroup$

First, the existence of limit at a point (say, $x_0$) does not depend on the value of the function at $x_0$. This is important because, without this idea, limit and continuity become more or less useless.

Now to show that $f$ has at most one limit as $x \to x_0$, we need to show that either the limit does not exist ("zero" limit), or the limit is unique if it exists ("one" limit).

In other words, we can assume that the limit exists, and let

$$ \lim_{x \to x_0} f(x) = l_1$$ and $$ \lim_{x \to x_0} f(x) = l_2$$. Now we want to show that $l_1 = l_2$. Start from the definition.

Given $\epsilon > 0$, $\exists \delta_1 > 0$ such that $$|f(x)-l_1| < \frac{\epsilon}{2}, \forall x \in V_{\delta_1}(x_0)\cap{S}\setminus{\{x_0 \}}$$ Also, $\exists \delta_2>0$ such that $$|f(x)-l_2|<\frac{\epsilon}{2}, \forall x \in V_{\delta_2}(x_0)\cap{S}\setminus{\{x_0\}}$$

. Now take $\delta = min\{\delta_1, \delta_2\}$.

$$|l_1-l_2| = |l_1-f(x)+f(x)-l_2| \le |l_1-f(x)|+|f(x)-l_2| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon, \forall x \in V_\delta(x_0) \cap S\setminus {\{x_0\}}$$. Since $x_0$ is a cluster point, $|l_1-l_2|$ is arbitrary small (but still non-negative) on a non-empty set. Can you conclude from here? (An extra hint: Suppose $\forall \epsilon > 0, 0\le a < \epsilon$, can you prove by contradiction that $a = 0$?)

$\endgroup$
0
$\begingroup$

Hint: A really classic way to approach uniqueness questions in analysis is by contradiction. Suppose $f$ had more than one limit (ie, two - say $a$ and $b$). What do you know about $\left| a-b \right|$? And how might this help you use the definition of a limit?

$\endgroup$

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.