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Consider the function $f : [1,2] \cup [3,4] \to \mathbb{R}$ where $f(x) = x$. Denote the domain by $D$. Is it continuous at $x=2$? This is not a homework question, even if it looks very fundamental and commonly asked. This is only an example with which I'm trying to demonstrate my lack of understanding.

  • Using the $\epsilon$ - $\delta$ definition : For any $\epsilon > 0$, there has to be a $\delta > 0$ such that for all $x \in D$, $0 \leq |x-2| < \delta \implies |f(x)-f(2)| < \epsilon$. Well simply set $\delta = \epsilon$ and we are done. We always say two sided limit has to exist, but this seems more like the existence of one sided limit at that point is enough.

  • Using the sequential definition : $\forall (x_n)_{n \in \mathbb{N}} \in D : \displaystyle\lim_{n \to \infty} x_n = 2 \implies \displaystyle\lim_{n \to \infty} = f(2)$. This too works, given $f(x) = x$, and the result is immediate. Again note that this doesn't require $(x_n)$ to be the sequences from the right side (in crude terms).

Can anyone explain what exactly I am missing?

oldsailorpopoye
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    You aren't missing anything. The technical definition of continuity accounts for one sided limits. Actually, continuity can be understood when the domain consists of isolated points only. – ProfOak Oct 06 '20 at 12:31
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    In the second bullet point, what did you mean by "$[1,2]$ is dense in $\mathbb{R}$"? – Jason DeVito - on hiatus Oct 06 '20 at 12:40
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    Does this help: https://math.stackexchange.com/questions/1914931/is-sqrt-x-continuous-at-0-because-it-is-not-defined-to-the-left-of-0? – Hans Lundmark Oct 06 '20 at 12:48
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    This notation I think is not appropriate: ${[1,2] \cup [3,4]}$ actually by $[a,b]$ we meant ${x| b\leq x\leq b}$ that is a priori a set. – C.F.G Oct 06 '20 at 13:23
  • @C.F.G The domain is a set too, right? A map is in fact a set (or more precisely, a subset of a set that we essentially call as the cartesian product set) with given conditions. I did mean $D = { x \in \mathbb{R} : x \in [1,2] \lor x \in [3,4]}$. – oldsailorpopoye Oct 06 '20 at 13:51
  • @JasonDeVito Thanks for pointing it out, it was certainly incorrect. I should have said subset of $\mathbb{R}$ (or has the properties of $\mathbb{R}$). – oldsailorpopoye Oct 06 '20 at 13:59
  • @others : Sure that does help. So even the topological definition of limit is domain dependent, and hence doesn't necessarily have to be two sided limit, right? The definitions convey that fact to me. – oldsailorpopoye Oct 06 '20 at 14:02
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    @charlesh: What C.F.G meant was that the domain is $[1,2]\cup [3,4]$, not ${[1,2]\cup [3,4]}$, the intervals are sets, so their union is also a set. Your domain has just a single element: the union $[1,2]\cup [3,4]$. Really, it's just the difference between ${x,y,z}$ and ${{x,y,z}}$ – Prasun Biswas Oct 06 '20 at 14:04
  • @PrasunBiswas Whoops! I'm too careless. I know what that means, I didn't see re-check the notation. Thanks for pointing it out again. – oldsailorpopoye Oct 06 '20 at 15:56

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