2
$\begingroup$

This might be a dumb question, but if so it will at least be easy to answer.

Let $X$ and $Y$ be metric spaces. Let $A \subseteq X$. Let $a \in A$. Let $f : A \to Y$ be a map. Then, define the phrase "$f$ is continuous at $a$'' to mean the following: $$ \forall \varepsilon > 0, \hspace{1mm} \exists\hspace{1mm} \delta > 0 \ni \forall x\in A, \hspace{1mm} d_X(x,a) < \delta \implies d_Y (f(x), f(a)) < \varepsilon $$ where $d_X$ and $d_Y$ denote the respective metrics of $X$ and $Y$.

Edit: We will apply this to a sequence $(a_n)$ in place of $f$, $\mathbb{Z}^+$ in place of $A$, and $n$ in place of $a$ (sorry for the confusion, the intent of the question refers to continuity of $(a_n)$ at $n$).

Consider any real valued sequence $(a_n)$, which is to say a map from $\mathbb{Z}^+$ to $\mathbb{R}$. Let $n$ be any positive integer. Let $a_n$ denote the value of the map $(a_n)$ at $n$. Let $\varepsilon > 0$. Choose $\delta = \frac{1}{2}.$ Then (using conventional metric functions, etc.), the set of $x\in\mathbb{Z}^+$ satisfying $d(x,n) < \delta$ is the singleton $\{n\}$. Moreover, we see that $x \in\{n\} \implies d(a_x, a_n) = 0 < \varepsilon$. Now, forget we fixed $\varepsilon$ and $n$, and we see that $(a_n)$ is continuous on $\mathbb{Z}^+.$

Is this wrong? Is it correct to say that $(a_n)$ is continuous at $n$? Must we preface this by saying that $(a_n)$ is continuous at $n$ as a function of $\mathbb{Z}^+,$ but not as a function of $\mathbb{R}$? Thanks/

$\endgroup$
  • $\begingroup$ Usual terminology - {$a_n$} has a limit. Alternatively {$a_n$} form a Cauchy sequence. $\endgroup$ – herb steinberg Jul 19 at 2:39
  • 1
    $\begingroup$ @herbsteinberg: That isn't the question, at all. In fact, every sequence (Cauchy or otherwise) is continuous. See my answer. $\endgroup$ – Cameron Buie Jul 19 at 2:48
  • 1
    $\begingroup$ I'm still trying to get through the fog of your question. It looks to me you have a sequence where all members are the same? $\endgroup$ – herb steinberg Jul 19 at 3:19
  • 1
    $\begingroup$ Not quite what you asked about, but: We can define a topology on $\Bbb N\cup \{\infty\}$ such that convergence of a sequence is equivalent to the ability to extend a sequence to a map that is continuous at $\infty$ ... $\endgroup$ – Hagen von Eitzen Jul 19 at 11:06
  • $\begingroup$ Hi everyone, sorry that I didn't update this until now, but I clarified the statement and, hopefully, the meaning of the question. That was my mistake. I am still curious if it is meaningful or appropriate to say that a sequence is continuous at every positive integer $n$ and/or discontinuous at real numbers $x$ satisfying $x \not\in\mathbb{Z}^+$. $\endgroup$ – Thomas Winckelman Jul 19 at 14:57
4
$\begingroup$

The answer (under the conventional metrics, etc.) is yes! Any sequence $\Bbb Z^+\to Y$ is continuous on its entire domain, or at every $a\in\Bbb Z^+.$ However, we'd typically say "at every $n\in\Bbb Z^+,$" for a few reasons. In this particular case, one will often denote the sequence of points $a_n$ by $a,$ rather than by $(a_n),$ so saying that such a sequence is "continuous at $a$" may seem nonsensical. Others will use $a$ to refer to the limit of such a sequence, if it exists, in which case it still wouldn't make sense. It's better not to equivocate on what we're calling our domain elements.

More generally, the set of positive integers under the usual metric is an example of what is known as a "discrete space." Given any discrete space $X$ and any topological space $Y,$ we have for all functions $f:X\to Y$ that $f$ is continuous on $X,$ since every subset of $X$ is open.

$\endgroup$
  • 1
    $\begingroup$ But the question is whether $(a_n)$ is continuous at $a$... $\endgroup$ – David C. Ullrich Jul 19 at 11:04
  • 1
    $\begingroup$ Good point, David. I had been connecting "$a$" to an arbitrary point of the domain, as I believe the OP intended, but I hadn't considered the potential for confusion. $\endgroup$ – Cameron Buie Jul 19 at 11:32
  • $\begingroup$ That was confusing, sorry, I tried to improve the question now. Also haha I have to admit I'm not knowledgeable on topology, so while your answer is probably helpful to others (so it's still great that you posted it), and I appreciate your willingness to share your knowledge, I personally lack the knowledge to find meaning in this answer... $\endgroup$ – Thomas Winckelman Jul 19 at 15:02
  • 1
    $\begingroup$ @ThomasWinckelman: Your question is clearer now. As for the topology bit, that's more of a side note that you may learn more about later. In the sense of metrics, the set $\bigl\{x\in A\mid d_X(x,a)<\delta\bigr\}$ is often referred to as "the open ball in $A$ centered at $a$ with radius $\delta.$" What you've shown is that in this particular example, when the radius is small enough, that open ball is a singleton containing only the center point, and consequently, the function value in that ball is constant. Therefore, the sequence is continuous at that point, as you suspected. (cont'd) $\endgroup$ – Cameron Buie Jul 20 at 14:43
  • 1
    $\begingroup$ ... The topological proof uses a similar idea, but without reference to a metric. To address your extension to the question in the comments above, we cannot say that the sequence is (dis)continuous at real numbers $x\notin\Bbb Z^+,$ because the sequence is not defined there. We can only talk about being (dis)continuous at points in a function's domain--that's what the "$a\in A$" condition means. $\endgroup$ – Cameron Buie Jul 20 at 14:46

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.