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Every open sphere in usual metric space Ru is an open interval. But the converse is not true; as
$(-\infty,+\infty)$ is an open interval in $\mathbb{R}$ but not an open sphere. My question is can a I write similar statement for $\mathbb{R}^2$, $\mathbb{R}^3$, and so on, taking Cartesian product of open intervals for corresponding space? Like $(-\infty,+\infty)\times (-\infty,+\infty)$; i.e their Cartesian product in $\mathbb{R}^2$ is not an open sphere? I guess I am not using mathematical words properly to explain my question. And I also want to know whether for only one such interval in Ru The statement has its converse not true? Is this statement has its converse not true only in Ru?

Empiricist
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Kavita
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  • In $\mathbb{R}^2$, Cartesian products of open intervals are sort of rectangles, not really sphere-like. – André Nicolas Oct 05 '15 at 02:24
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    The open interval $(-n,n)$ is an open ball, not a sphere. A sphere is the boundary of a closed ball. – William Stagner Oct 05 '15 at 02:28
  • Yes, I have seen such statement that (-infinity, +infinity) is not an open sphere in R with usual. – Kavita Oct 05 '15 at 02:29
  • There is no term "open sphere". A sphere is just the set of points in $\mathbb R^n$ that are a fixed distance from $0$. A sphere in $\mathbb R$ is just a pair of points ${-n, n}$. – William Stagner Oct 05 '15 at 02:33
  • But still I am not clear how we get open ball or open sphere in R? I mean in R I just know that every open sphere is open interval. What about open ball? Not getting the difference. – Kavita Oct 05 '15 at 02:35
  • A sphere is just the boundary of a ball. You mean an open ball to include the interior. For example, think about a circle vs a disk. A circle is just a curve and doesn't include the bounded region. – Michael Burr Oct 05 '15 at 02:50

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