I'm self-studying from the book Understanding Analysis by Stephen Abbott, and I have the feeling that the author is being careless about limit points in his theorems or I am not understanding something (probably the latter). For instance, for the following theorem:

(Algebraic Continuity Theorem). Assume $f: A \to \mathbb{R}$ and $g: A \to \mathbb{R}$ are continuous at a point $c \in A$. Then:

  1. $kf(x)$ is continuous at $c$ for all $k \in \mathbb{R}$;
  2. $f(x) + g(x)$ is continuous at $c$;
  3. $f(x)g(x)$ is continuous at $c$; and
  4. $f(x)/g(x)$ is continuous at $c$, provided the quotient is defined.

Proof. All of these statements can be quickly derived from Theorem$^1$ and Theorem$^2$ (see below for the two theorems).

The only way I am able to prove the above theorem is by using theorem$^2$ part 2. However, part 2 is only true if the point $c \in A$ is a limit point. Since, for the above theorem, this is not a restriction on $c$, I don't understand how to prove it. Any help is much appreciated.

As an extra question: as far as I understand part 1 and part 3 of theorem$^2$ hold also true for continuous functions at $c$ if $c$ is not a limit point, right? What about part 4 of theorem$^2$? Does $c$ is need to be a limit point, or is this not necessary?

$^1$ (Algebraic Limit Theorem for Functional Limits). Let $f$ and $g$ be functions defined on a domain $A \subseteq \mathbb{R}$, and assume $\lim_{x\to c} f(x) = L$ and $\lim_{x \to c}g(x) = M$ for some limit point $c$ of $A$. Then:

  1. $ \lim_{x \to c} kf(x) = kL$ for all $k \in \mathbb{R}$,
  2. $\lim_{x \to c} [ f(x) + g(x)] = L + M$,
  3. $\lim_{x \to c} [f(x)g(x)] = LM$, and
  4. $\lim_{x \to c} f(x)/g(x) = L/M$, provided $M \neq 0$.

$^2$ (Characterizations of Continuity). Let $f : A \to \mathbb{R}$ and $c \in A$ be a limit point of $A$ [emphasis mine]. The function $f$ is continuous at $c$ if, and only if, any one of the following conditions is met:

  1. For all $\epsilon > 0$, there exists a $\delta > 0$ such that $|x-c| < \delta$ (and $x \in A$) implies $|f(x) - f(c)| < \epsilon$;
  2. $\lim_{x \to c} f(x) = f(c)$;
  3. For all $V_\epsilon (f(c))$, there exists a neighborhood $V_\delta (c)$ with the property that $x \in V_\delta(c)$ (and $ x \in A$) implies $f(x) \in V_\epsilon(f(c))$;
  4. If $(x_n) \to c$ (with $x_n \in A$), then $f(x_n) \to f(c)$.

For completeness, I will also write down the way the author defines functional limits and continuity:

Definition$^3$. Let $f: A \to \mathbb{R}$, and let $c$ be a limit point of the domain $A$. We say that $\lim_{x \to c} f(x) = L$ provided that, for all $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $0 < |x-c| < \delta$ (and $x \in A$) it follows that $|f(x)-L| < \epsilon$.

Definition$^4$. A function $f: A \to \mathbb{R}$ is continuous at a point $c \in A$ if, for all $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $|x-c| < \delta$ (and $x \in A$) it follows that $|f(x)-f(c)| < \epsilon$.

  • 1
    $\begingroup$ It seems to me that the definition of continuity is indeed slightly incorrect (is it a definition, or just a theorem about continuity at a point?). Note that none of 1-4 in the "Characterizations of continuity" actually requires $c$ to be a limit point of $A$. If it is not, i.e. if $c$ is an isolated point of $A$, then there is no way for $f$ to be discontinuous at $c$. $\endgroup$ Aug 31, 2014 at 14:57
  • $\begingroup$ @MarcinŁoś thanks for your reply. What I had written were theorems. I've now also written the definitions of functional limits$^3$ and continuity$^4$ as the author has written them. According to the author, $c$ must be a limit point for the definition of the functional limit, but $c$ does not need to be a limit point for the definition of continuity of a function. According to these definition, I believe $c$ must be a limit in order for part 2 in "Characterizations of continuity" to be valid. Do you agree with that, or am I wrong? $\endgroup$
    – Hunter
    Aug 31, 2014 at 15:12
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    $\begingroup$ Yes - as it's written, definition of the limit clearly requires $c$ to be a limit point of $A$. So, yes - "Characterizations of continuity" needs $c$ to be a limit point of $A$, so everything seems correct now. +1 for careful reading and urge to understand the text fully :) $\endgroup$ Aug 31, 2014 at 16:33
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    $\begingroup$ No, the Algebraic Continuity Theorem is valid regardless of $c$. What is ... well, valid but a bit misleading is the statement that we can use Algebraic Limit Theorem for Functional Limits to prove it. We need to consider two cases - if $c$ is a limit point of $A$, we can use the Algebraic Limit Theorem. Otherwise, these functions are continuous at $c$ for example by point 4 from Characterizations of Continuity - the only sequence in $A$ convergent to $c$ is a constant one, so clearly $f(x_n)\rightarrow f(x)$. $\endgroup$ Aug 31, 2014 at 20:35
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    $\begingroup$ By all means, do so - answering your own questions is perfectly fine. Should you have any trouble, feel free to ask, I should be here for like next 3 hours. $\endgroup$ Aug 31, 2014 at 22:04

1 Answer 1


I will try to prove the Algebraic Continuity Theorem without using part 2 of Characterizations of Continuity; if I succeed, then I can conclude that for the Algebraic Continuity Theorem the point $c \in A$ does not need to be a limit point.

  1. In order to prove that $kf(x)$ is continuous at $c$, we need to find $\delta > 0$ such that: \begin{equation} |x-c| < \delta \implies |kf(x) - k f(c)| = |k||f(x)-f(c)| < \epsilon \end{equation} But since $f$ is assumed to be continuous, we can choose $\delta$ such that: \begin{equation} |x-c| < \delta \implies |f(x)-f(c)| < \frac{\epsilon}{|k|} \end{equation} and consequently: \begin{equation} |x-c| < \delta \implies |k||f(x)-f(c)| < |k| \frac{\epsilon}{|k|} = \epsilon \end{equation} as desired.
  2. We now need to prove that there is a $\delta > 0$ such that: \begin{equation} |x-c| < \delta \implies |f(x) + g(x)- (f(c)+ g(c))| < \epsilon \end{equation} Since $f$ is assumed to be continuous, we can choose a $\delta_1 > 0$ such that: \begin{equation} |x-c| < \delta_1 \implies |f(x)-f(c)| < \frac{\epsilon}{2} \end{equation} and since $g$ is assumed to be continuous, we can choose a $\delta_2 > 0$ such that: \begin{equation} |x-c| < \delta_2 \implies |g(x)-g(c)| < \frac{\epsilon}{2} \end{equation} Let us now set $\delta = \mathrm{min} \{ \delta_1, \delta_2 \}$. Then, using the triangle inequality, we can write: \begin{equation} \begin{aligned} |x-c| < \delta \implies |f(x) + g(x)- (f(c)+ g(c))|&=|(f(x)-f(c)) + (g(x) - g(c))| \\& \leq |f(x)-f(c)| + |g(x) - g(c)| \\& < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon \end{aligned} \end{equation}
  3. Next, we need to prove that there exist a $\delta >0$ such that: \begin{equation} |x-c| < \delta \implies |f(x)g(x) - f(c)g(c) | < \epsilon \end{equation} Note that: \begin{equation} \begin{aligned} |f(x)g(x) - f(c)g(c)| & = |f(x)g(x) - f(c)g(x) + f(c)g(x) - f(c)g(c)| \\& \leq |f(x)g(x) - f(c)g(x)| + |f(c)g(x) - f(c)g(c)| \\& =|g(x)||f(x) - f(c)| + |f(c)||g(x) - g(c)| \end{aligned} \end{equation} Now, choose $\delta_1 > 0$ such that: \begin{equation} |x-c| < \delta_1 \implies |f(x)-f(c)| < \frac{1}{|M|} \frac{\epsilon}{2} \end{equation} where $|g(x)| \leq |M|$ for all $ x \in A$. Furthermore, choose $\delta_2 > 0$ such that: \begin{equation} |x-c| < \delta_2 \implies |g(x)-g(c)| < \frac{1}{|f(c)|} \frac{\epsilon}{2} \end{equation} Let us now set $\delta = \mathrm{min} \{ \delta_1, \delta_2 \}$. Then: \begin{equation} \begin{aligned} |f(x)g(x) - f(c)g(c)| & \leq |g(x)||f(x) - f(c)| + |f(c)||g(x) - g(c)| \\& \leq |M||f(x) - f(c)| + |f(c)||g(x) - g(c)| \\& < |M|\frac{1}{|M|} \frac{\epsilon}{2} + |f(c)|\frac{1}{|f(c)|} \frac{\epsilon}{2} = \epsilon \end{aligned} \end{equation} as desired.
  4. In order to prove this statement, we need to show that $1/g(x)$ is continuous. Thus, we need to show that there exists a $\delta >0$ such that: \begin{equation} |x-c| < \delta \implies \left| \frac{1}{g(x)} - \frac{1}{g(c)} \right| < \epsilon \end{equation} Note that: \begin{equation} \left| \frac{1}{g(x)} - \frac{1}{g(c)} \right| = \frac{1}{|g(x) g(c)|} |g(x)- g(c)| \end{equation} Since $g(x)$ is continuous, we can choose $\delta > 0$ such that: \begin{equation} |g(x) - g(c)| < |K| |g(c)| \epsilon \end{equation} where: \begin{equation} \frac{1}{|g(x)|} \leq \frac{1}{|K|} \end{equation} for all $x \in A$. Therefore: \begin{equation} \begin{aligned} \left| \frac{1}{g(x)} - \frac{1}{g(c)} \right| & \leq \frac{1}{|K g(c)|} |g(x)- g(c)| \\& < \frac{1}{|K g(c)|} |K| |g(c)| \epsilon = \epsilon \end{aligned} \end{equation} as desired.
  • 1
    $\begingroup$ The proofs are essentially correct. I believe you should explicitly address potential issues with zeros in the denominator, especially in 4 (if $g(c)\neq 0$, then, by continuity also $g(x)\neq 0$ in some neighborhood of $c$), but otherwise it's a solid, rigorous proof. Note that you have essentially proven the Algebraic Limit Theorem here (slightly extended, since the definition of limit in this book requires $c$ to be a limit point of $A$). Note also that you could have used the Algebraic Limit Theorem instead of reproving it here, if you wanted to. $\endgroup$ Sep 1, 2014 at 0:55
  • 1
    $\begingroup$ The argument goes as follows: If $c$ is a limit point of $A$, then by the Algebraic Limit Theorem, limit of $kf(x)$, $f(x)+g(x)$ etc. as $x\rightarrow c$ exists and is equal to $kf(c)$, $f(c)+g(c)$ etc., hence by (2) in Characterizations of Continuity, all 4 functions are continuous at $c$. $\endgroup$ Sep 1, 2014 at 0:55
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    $\begingroup$ If, on the other hand, $c$ is not a limit point of $A$, then there exists $\delta > 0$ such that $V_\delta(c)$ does not contain any points of $A$ besides $c$. Hence, (3) in Characterizations of Continuity is trivially satisfied - for any $V_\epsilon(f(c))$ our $V_\delta(c)$ satisfies the condition, as the only $x\in V_\delta(c)\cap A$ is $c$, and $f(c)\in V_\epsilon(f(c))$ for any $\epsilon$. $\endgroup$ Sep 1, 2014 at 0:56
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    $\begingroup$ As it's 3 AM here, I'll be going now ;) If you have any further questions, I'll be happy to answer tomorrow. Good luck! $\endgroup$ Sep 1, 2014 at 1:00
  • $\begingroup$ @MarcinŁoś ahh ok, that is indeed an easier way. Thanks so much! If you want to, you can collect your comments and post it as an answer, and then I will happily accept it. Good night ;) $\endgroup$
    – Hunter
    Sep 1, 2014 at 1:30

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