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Given a function $f$; $$ f(x) = \frac{x}{1+x^2} $$

the support of $f$ is $$ \mbox{supp}f=\begin{cases} \overline{\{(-\infty, 0), (0, \infty)\}}=(-\infty, \infty)=\mathbb{R}&&(1)\\ \overline{(-\infty, 0) \cup (0, \infty)}=(-\infty, \infty)=\mathbb{R}&&(2)\\ \end{cases} $$

Which one is correct?

Is interval itself already a set? or do I have to write bracket; '{' and '}'?

Danny_Kim
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1 Answers1

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$(2)$ is correct, the first is a set containing a single set (which contains all reals except $0$). There is no obvious meaning for the overbar above that set either. (closure? relative to what topology?)

Justin Benfield
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  • Ah, overbar means closure in my notation. I wrongly thought it is a common sign like +, -. Thank you :) – Danny_Kim Mar 12 '16 at 09:06
  • It is common notation, but the very concept of closure is topological in nature, hence the set in question must be a (sub)-set of a topological space in order for that notation to even have meaning. – Justin Benfield Mar 12 '16 at 09:07
  • Sorry, I do not understand your explanation. You mean overbar has onother meaning? – Danny_Kim Mar 12 '16 at 09:08
  • The very idea of a set having a closure is dependent upon the topology of the space that set came from. In the present situation you have given no indication what topological space the first set came from hence there is no obvious way to interpret what that set's closure is (the 2nd is obviously a subset of $\mathbb{R}$ hence the standard topology on $\mathbb{R}$ is assumed). – Justin Benfield Mar 12 '16 at 09:12
  • Hmm, I've never learned about topology. That's why I do not understand of your answer. (Another reason is my poor English. T.T) I understood that second one can be guessed as $\mathbb{R}$, however first one is not. Therefore, I have to specify the specific topology, right? Although I don't know what the topology is, I felt it has many types according to your answer. – Danny_Kim Mar 12 '16 at 09:18
  • The first one also isn't equal to the set to it's right. Given a set $X$ a topology $\mathfrak{T}$ for the set $X$ is a set of subsets of $X$, such that:
    1. ${ }$ and $X$ are in $\mathfrak{T}$.
    2. Any union of sets in $\mathfrak{T}$ is again in $\mathfrak{T}$.
    3. Any finite intersection of sets in $\mathfrak{T}$ is again in $\mathfrak{T}$.

    The sets in $\mathfrak{T}$ are declared to be open in the topological space $(X,\mathfrak{T})$. A set is closed in $X$, with respect to $\mathfrak{T}$, if its complement in $X$ is an open set (hence in $\mathfrak{T}$).

     – Justin Benfield Mar 12 '16 at 09:30
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    @JustinBenfield Just a small correction of your answer. (1) is incorrect, because there is no obvious meaning for the closure of such set. However, without the bar, the set ${(-\infty,0),(0,+\infty)}$ is a set with two elements (and its elements are intervals in $\mathbb{R}$. It is not true that the set described in (1) is a single set containing all reals except $0$. – Darío G Mar 12 '16 at 09:48
  • I am impressed with so kind and good comments! – Danny_Kim Mar 12 '16 at 09:48
  • @user27454 You're right, missed the comma and assumed there was a union there just as with the set below. Either way, it's not the set that the OP intended to describe. – Justin Benfield Mar 12 '16 at 09:50
  • @JustinBenfield If you replace the comma by an union symbol, then you get ${(-\infty,0)\cup (0,\infty)}$ which is a set with only one element and not a set containing all reals except $0$. Perhaps is it what you meant when you wrote it was a "single" set? – Darío G Mar 12 '16 at 10:20
  • @user27454 Yes, I meant that it was a set containing a single element, namely another set; and that other set is all reals except $0$. – Justin Benfield Mar 12 '16 at 10:22