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I was always under the impression that in order to take a limit, I need to have a metric defined on my underlying space.

For $f:\mathbb{R}\to \mathbb{R}$, $ \lim_{x\to x_0}f(x)=a $ means that for all $\epsilon >0 $ there exists a $\delta(\epsilon)>0$ such that $|f(x)-a|<\epsilon$ whenever $0< |x-x_0|<\delta$. The notion of a limit uses the underlying metric $|\cdot|$ of $\mathbb{R}$.

Is there a consistent way of dispensing with the metric and still define a limit operation in some topological space? Are nets able to do this?

Antonio Maria Di Mauro
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EEEB
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3 Answers3

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The notion of limit is well-defined for any topological space, even non-metric ones.

Here is the correct definition: let $f : X \rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x \in X$ is the point $y \in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M\setminus\{x\}) \subset N$.

But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) \rightarrow x$, one has $f(x_n) \rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.

NDB
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TheSilverDoe
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    Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces. – chi Mar 17 '19 at 19:05
  • @chi Yes, thank you for this comment ! Indeed the precision can be usefull. – TheSilverDoe Mar 17 '19 at 19:07
  • Not just funny for non-Hausdorff spaces, for example if we give $\Bbb Z$ the discrete topology then $\lim_{x\to a}f(x)=L$ holds for every $f:\Bbb Z\to Y$, every $L\in Y$ and every $a\in \Bbb Z$. – David C. Ullrich Sep 05 '22 at 14:55
  • This is incorrect. Take $f:\mathbb{R}\to\mathbb{R}$ with $f(0)=1$ and $f(x)=3$ for $x\neq0$. Then, take the ball of center $3$ and radius $1$ as $N$, then there is no neighborhood $M$ of $0$ such that $f(M)\subset N$. This is because $0\in M$ implies that $1\in f(M)$, but $1\notin N$. On the other hand, we would like $\lim_{x\to0}f(x)=3$. This is defining continuity at a point. – NDB Aug 02 '23 at 19:33
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    @NDB But the condition is that there exist a neighborhood $M$ of $x$ such that $f(M \setminus {x}) \subset N$, not such that $f(M) \subset N$. – gf.c Jan 14 '24 at 20:21
  • $x$ should be a limit point of $X,$ not an element of it. – GraphicsMuncher Mar 29 '24 at 04:53
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A net is a function from an directed set $(I, \le)$ (say) to a space $X$.

$f: I \to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 \in I$ (depending on $O$, in general) such that for all $i \in I, i \ge i_0$ we know that $f(i) \in O$.

The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=\mathbb{N}, \le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,\varepsilon), \varepsilon>0$ as these form a local base at $x$.

I think that $\lim_{x \to a} f(x)$ can be defined by considering all nets $n$ on $X\setminus \{a\}$ that converge to $a$, and if all those nets have the property that $f \circ n$ is a net in $Y$ converging to the same $b \in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.

If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.

Henno Brandsma
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  • Hi professor, I read your answer and I'm thinking if to say that $\text{lim}_{x\rightarrow a}f(x)=b$ is really necessary that any net $n$ on $X\setminus{a}$ that converge to $a$ is such that $f\circ n$ is a net converging to the same $b$. Indeed I would say that the function $f$ has a limit $b$ iff for some $a\in X$ there exist some net $n$ on $X\setminus{a}$ that converge to $a$ and that has the property that $f\circ n$ is a net in $Y$ converging to $b$. Perhaps this is a bad definition? What can you say about? – Antonio Maria Di Mauro Jun 28 '20 at 10:11
  • You want to generalise the idea that limits from the left and the right (in the reals) are the same, or limits from all directions in the plane. So that would be a bad definition. @AntonioMariaDiMauro – Henno Brandsma Jun 28 '20 at 11:31
  • Okay. However for metric spaces the uniqueness of the limit could be it proved using some results about hausdorff or normal spaces? For example in a hausdorff space any sequence (obviously a sequence is a net) converges at most one point. What can you say about? – Antonio Maria Di Mauro Jun 28 '20 at 12:38
  • @AntonioMariaDiMauro unicity of limits will be provable in Hausdorff spaces. And in a Hausdorff space of course a sequence will not converge to at least one point, that is nonsense. What do you really mean? – Henno Brandsma Jun 28 '20 at 12:40
  • I want say at most, excuse me for mistake. – Antonio Maria Di Mauro Jun 28 '20 at 12:41
  • @AntonioMariaDiMauro It's the same for nets, clearly. That's what I mean by "limits are unique". – Henno Brandsma Jun 28 '20 at 12:42
  • Okay, so why you say that $\text{lim}{x\rightarrow a}f(x)$ iff all nets $n$ on $X\setminus{A}$ that converge to $a$ have the property that the nets $f\circ n$ converge to the same $b\in Y$? Infact for what now we said it seems to me it could say that $\text{lim}{x\rightarrow a}f(x)=b$ iff there exist a net $n$ on $X\setminus{A}$ that converge to $a$ such that the net $f\circ n$ converge to the $b\in Y$, so what is my mistake? – Antonio Maria Di Mauro Jun 28 '20 at 12:47
  • Perhaps could it be that there exist $x_\lambda$ and $y_\lambda$ such that $f(x_\lambda)\rightarrow a$ and $f(y_\lambda)\rightarrow b$ and $a\neq b$, because the results for hausdorff spaces guarantee that the limits $a$ and $b$ are unique for $f(x_\lambda)$ and $f(y_\lambda)$ but we don't know if $a\neq b$, is this what you want say? – Antonio Maria Di Mauro Jun 28 '20 at 12:50
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    @AntonioMariaDiMauro In your alternative definition $\lim_{x \to 0} \sin(\frac{1}{x}) = 1$ would hold. – Henno Brandsma Jun 28 '20 at 12:50
  • Okay, so the limit of a function is unique for definition and not for a theorem, right? – Antonio Maria Di Mauro Jun 28 '20 at 12:52
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    @AntonioMariaDiMauro The definition is for "$\text{lim}(f,a,b):= \lim_{x \to a} f(x)=b$" where $f$,and $a,b$ are fixed. A theorem then says that $\text{lim}(f,a,b) \land \text{lim}(f,a,b') \to b =b'$ when $Y$ is a Hausdorff space. Continuity of $f$ is irrelevant. – Henno Brandsma Jun 28 '20 at 12:55
  • Okay, so by your definition if $x_\lambda$ and $y_\lambda$ are net that converges to $x_0$ then if $f(x_\lambda)\rightarrow a$ and $f(y_\lambda)\rightarrow b$ and if $\text{lim}{x\rightarrow x_0}f(x)$ exist then by definition of limit $f(x\lambda)\rightarrow b$ and $f(y_\lambda)\rightarrow a$ and generally it could be $a\neq b$ but for hausdorff space it must be $a=b$, right? did I have understand? – Antonio Maria Di Mauro Jun 28 '20 at 13:01
  • @AntonioMariaDiMauro no, you misquote me.. – Henno Brandsma Jun 28 '20 at 13:05
  • Excuse me, forgive my confusion. What I did't understand? Could you explain to me, please? – Antonio Maria Di Mauro Jun 28 '20 at 13:06
  • Let us continue this discussion in chat. – Henno Brandsma Jun 28 '20 at 13:07
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The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X \to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) \subset V$.

Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_{i \in I}$ converges to some point $x$ we have $x$ we have $(f(x_i))_{i \in I}$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_{i \in I}$ converges to $x$ we have $x$ we have $(f(x_i))_{i \in I}$ converges to $f(x)$.

Kavi Rama Murthy
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