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The definition of the limit for a function $f:\Omega\rightarrow{}Y$, where the image $Y$ is a topological space and the domain $\Omega$ is a subset of some other topological space $X$, is that $$\lim_{x\rightarrow{}x_0}f(x)=y_0$$ if for all neighborhoods $V$ of $y_0$ there exists a neighborhood $U$ of $x_0$ such that $f(U\cap\Omega-\{x_0\})\subseteq{}V$. Why do we exclude $x_0$ from $U\cap\Omega$? Are we trying to avoid something?

I can see that if there exists a neighborhood $U$ of $x_0$ such that $U\cap\Omega=\emptyset$ or $U=\{x_0\}$, then the limit as $x$ approaches $x_0$ of $f(x)$ could be any $y\in{}Y$. So under this definition the limit is trivial for points not in the closure of $\Omega$ or for isolated points. Why would we only want limit points to have non-trivial limits?

Also, considering that $Y$ is not a Hausdorff space, wouldn't $f$ be automatically continuous for isolated points? It would be continuous for every non-limit point, except that non-limit points that are not isolated points are not in the domain, which is a necessary condition for continuity at a point.

Does that also mean that if $\Omega$ has an isolated point $p$ then it can't be a Hausdorff space under the subspace topology inherited from $X$. Since every limit as $x$ approaches $p$ would have more than one value?

Martin Sleziak
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awesomeusername
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1 Answers1

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We exclude $x_0$ because the limiting behavior at $x_0$ does not depend on the definition of $f$ at $x_0$. If you included $x_0$, then functions continuous away from $x_0$ with a jump at $x_0$ would not have a limit there-but they should.

This gives the strange behavior you describe when $x_0\in \Omega$ is isolated-but if we're consistent in thinking that a limit of a function is about its behavior near a point, not at the point, then this is irrelevant.

Every function is continuous at an isolated point, regardless of Hausdorffness of the range. $\Omega$ is not a subspace of $Y$.

Your last paragraph doesn't make sense: if you're inducing a topology from $Y$, this can't be affected by its pre-existing topology, including limit points. Certainly $\Omega$ could have isolated points but have Hausdorff induced topology: consider the identity map from the discrete topology to a Hausdorff topology on the same set.

Kevin Carlson
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  • Sorry about my last paragraph, now I'm so confused about it that even I don't know what I was talking about there.. Anyway, could you expand on that argument about continuity at a jump at $x_0$? Where would the function not have a limit but it should? At the jump? – awesomeusername Nov 13 '14 at 01:42
  • Right, at the jump. Just think of a function on $\mathbb{R}$ zero at every point but $0$ and $1$ at zero. The limit of this function as $x\to 0$ is zero, but if you tried to include $0$ in the neighborhoood-definition, there would be no limit, because no whole neighborhood of zero is contained in any sufficiently small interval. – Kevin Carlson Nov 13 '14 at 07:14
  • I think I got it! Since the topology on the image would be ${{0}\cup{1},\emptyset,{0},{1}}$ there is no neighborhood $U$ of $0$ such that $f(U)\subseteq{0}$ or $f(U)\subseteq{1}$ so we exclude $0$ so that $f(U-{0})={0}\subseteq{0}$. Right? – awesomeusername Nov 13 '14 at 11:16
  • That's basically right, but you generally want to talk about neighborhoods in the range, rather than in the image: so you'd use $(-\epsilon,\epsilon)$ in the argument where you take ${0}$. – Kevin Carlson Nov 13 '14 at 15:42
  • I don't think it matters if I consider the topology of the codomain or the subspace topology of the image. Because if a point $y_0$ of the image has a neighborhood $V$ with respect to the codomain such that there doesn't exist a neighborhood $U$ of $x_0$ such that $f(U\cap\Omega-{x_0})\subseteq{}V$, then definitely there doesn't exist a neighborhood $U$ of $x_0$ such that $f(U\cap\Omega-{x_0})\subseteq{}V\cap{}Im(f)$. Right? – awesomeusername Nov 13 '14 at 17:11
  • Yes, these are equivalent-I've just never seen someone argue this way, and so people may do a double take if you choose to. You also run into trouble once you want something more interesting than continuity, e.g. smoothness. – Kevin Carlson Nov 13 '14 at 22:03
  • Ohh, ok! Thanks for the heads up! Also, there's something else I've been wondering that might be related.. I remember reading somewhere that some limits that exist in $\mathbb{R}^2$ don't exist in $\mathbb{C}$ or vice-versa, but aren't the topologies in those spaces the same? How can the limits not coincide? – awesomeusername Nov 13 '14 at 22:09
  • Right, as you notice should be the case, the limits are exxactly the same-most likely you're slightly misremembering. What is true is that differentiability on $\mathbb{C}$ is a much stronger notion than differentiability on $\mathbb{R}$. But this isn't really an issue of topology at all, except in its rich consequences. – Kevin Carlson Nov 13 '14 at 22:23
  • How can differentiability in $\mathbb{C}$ and $\mathbb{R}^2$ be different? Isn't differentiability stated in terms of limits which in turn are defined in terms of the topologies on those spaces? Do people mean that holomorphicity is a much stronger notion than differentiability for a point in $\mathbb{C}$? Is it a linguistic issue that sometimes holomorphicity is synonymous with "complex differentiability"? – awesomeusername Nov 14 '14 at 12:28
  • Yes, I mean to bring up holomorphicity. – Kevin Carlson Nov 14 '14 at 16:48