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I'm working on a question that wants me to write down the interior points of an interval contained in a metric space.

$Let X=((1,7],d_{E})$ be a subspace of the metric space $(\mathbb{R},d_{E})$. Let $A=[5,7]$. Find the interior points of A regarded as a subset of

  • $X$
  • $(\mathbb{R},d_{e})$

My answer is: (5,7) for both. The question however also states that they are not the same. How so? What am i misunderstanding?

Regards

Adeeb
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2 Answers2

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This is because, in the first case, the interior points are $(5,7]$.

Indeed, $(5,7]=(5,8)\cap (1,7]$, so it is open in $X$.

And clearly $5$ is not in the interior of $[5,7]$, since every open neighborhood contains points smaller that $5$.

Note: see here for the notion of subspace topology, to see in particular why I claim that $(5,7]$ is open in $X$: http://en.wikipedia.org/wiki/Subspace_topology

Alternative: write $$ (5,7]=\{x\in X\;|\; d_E(x,6.5)<1.5\} $$ to see that this is open in $X$, as an open ball.

Julien
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  • I'm havin' difficulty understanding why 7 is included. Why did you write this $(5,7]=(5,8)\cap (1,7]$ ? – Adeeb Feb 17 '13 at 14:31
  • Read the link on subspace topology and let mw know if you still have questions. – Julien Feb 17 '13 at 14:32
  • @Adeeb I've added an alternative viewpoint. – Julien Feb 17 '13 at 14:35
  • @Adeeb The example added to Julien's answer IS an open ball in $X$ that contains $7$. – Tom Oldfield Feb 17 '13 at 14:52
  • Ok, now I fully understand. How would you define the interior points of a set which is not even known in the metric space? for instance; interior points of $A=(9,12]$ in $V=((1,7],d_E)$, is it the empty set? – Adeeb Feb 17 '13 at 16:40
  • @Adeeb Well, $A$ has empty intersection with $V$ so it does not really make sense. But yes, the interior of $\emptyset$ is $\emptyset$. – Julien Feb 17 '13 at 16:41
  • I was looking for this " A has empty intersection with V so it does not really make sense". I was not referring to the interior of $\emptyset$ – Adeeb Feb 18 '13 at 20:45
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The interior of a set, $\operatorname{Int}(A)$ is usually defined to be the union of all open subsets contained in the set $A$.

In this case, if you work with the induced subspace topology on $X$ then, for example, $(6,8)$ is open in $\mathbb{R}$. Hence $(6,8)\cap X = (6,7]$ is open in $X$. Hence $7$ is in the interior of $(5,7]$.

Even if you prefer working with the just $(X,d_E)$ as a metric space in it's own right and not as a subspace, then it's still true that $(6,7]$ is open in $X$ since it is the open ball around $7$ of radius 1.

I hope that's clear now?

Tom Oldfield
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  • I understand from Real Analysis that interior points are points which are not on the boundary of a set. How can 7 be included as an interior point of A? – Adeeb Feb 17 '13 at 14:26
  • @Adeeb But then how do you define the boundary? I think it's usual to define the interior of a set as above, then define the closure of a set $A$ as being the intersection of all closed sets containing $A$. The boundary is then everything in the closure not in the interior. The idea of a boundary is that it "separates the inside and the outside". This explains the difference between the interior points here. In $\mathbb{R}$, $7$ separates $A$ from part of the real line, so is on the boundary. In $X$ though, $7$ doesn't separate $A$ from anything, so can't be said to be on the boundary. – Tom Oldfield Feb 17 '13 at 14:32