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I just have a quick question, as stated in the title.

Is $\mathbb{R}^2$ boundary-less?

Thank you very much. :-)

1LiterTears
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    Isn't each topological space boundary-less? – Stefan Hamcke Jun 11 '13 at 19:31
  • That is good to know, thanks Stefan H.... But you mean the whole set, not topological subset, right? As @Shaun Ault answered below, $\mathbb{R}^2$ is boundaryless if it sits in $\mathbb{R}^2$, but is with boundary it it sits in $\mathbb{R}^n$. – 1LiterTears Jun 11 '13 at 19:43
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    That's why you should always specify the underlying space. $A$ may have a boundary in $X$, but $A$ regarded as a subset of the space $(A,\tau)$ (where $\tau$ can be any topology, but mostly one takes the subspace topology) has no boundary. – Stefan Hamcke Jun 11 '13 at 20:24

2 Answers2

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That depends on whether you consider $\mathbb{R}^2 \subseteq \mathbb{R}^2$ or $\mathbb{R}^2 \subseteq \mathbb{R}^n$ for $n \geq 3$. In the former case, there are no points in the boundary; in the latter, $\partial \mathbb{R}^2 = \mathbb{R}^2$.

Shaun Ault
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  • Fairly clear, thank you very much!! – 1LiterTears Jun 11 '13 at 19:44
  • This question arises from the settlement: http://math.stackexchange.com/questions/417789/a-direct-application-of-sards-theorem. So to my understanding, it is boundaryless under this scenario. Thanks. – 1LiterTears Jun 11 '13 at 19:49
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    Ah... boundary can mean something different in manifolds. $\mathbb{R}^2$ is boundaryless as a manifold, because all points can be surrounded by an open 2-disk. – Shaun Ault Jun 11 '13 at 21:16
  • For $X \in \mathbb{R}^n$ in the transversality theorem - it is still boundaryless right? (Suppose that $F: X \times S \rightarrow Y$ is a smooth map of manifolds, where only $X$ has boundary, and let $Z$ be any boundaryless submanifold of $Y$. If both $F$ and $\partial F$ are transversal to $Z$, then for almost every $s \in S$, both $f_s$ and $\partial f_s$ are transeversal to $Z$.) – 1LiterTears Jun 12 '13 at 17:46
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    For the purposes of this theorem, $\mathbb{R}^2$ is a manifold without boundary. The following link may help, though you should also consult a text in which manifolds with and without boundary are defined precisely. http://en.wikipedia.org/wiki/Manifold#Manifold_with_boundary – Shaun Ault Jun 13 '13 at 01:13
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Hint $$\partial A= \overline{A}\setminus \overset{\circ}{A}$$ so what's the adherence and the interior of $\mathbb R^2$?

  • Hi thank you for your help. Could you please inform me that what do you mean by $\overset{\circ}{A}$? Thank you very much! – 1LiterTears Jun 11 '13 at 19:29
  • Also, by adherence, do you mean closure? Thanks. – 1LiterTears Jun 11 '13 at 19:29
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    @MathSnail The circle above a topological space denotes the interior. And I believe you are correct that adherence means closure in this context. – Potato Jun 11 '13 at 19:35
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    Yes the adherence is the closure. –  Jun 11 '13 at 19:36
  • Hi @SamiBenRomdhane, Thanks very much for your guidance. So the adherence of $\mathbb{R}^2$ is $\mathbb{R}^2$, but the interior of $\mathbb{R}^2$ is also $\mathbb{R}^2$. So $\mathbb{R}^2$ is boundaryless. – 1LiterTears Jun 11 '13 at 19:45
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    Yes this is the right answer. –  Jun 11 '13 at 19:54