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Does the point $x$ belong to the $\epsilon$-neighbourhood of $x$? According to the definition, the neighbourhood of $x$ consists of all $y$ such that $d(y,x)< \epsilon$. Does $x$ belong to the neighbourhood?

Micah
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Parul
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2 Answers2

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Note that $x \in $ the $\epsilon$-neighborhood of $x$, by your definition, because

$$d(x, x) = 0 \lt \epsilon\;\text{ since }\epsilon> 0$$

amWhy
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By definition of a metric $d(x,y) = 0$ iff $x=y$. Hence for every $\epsilon > 0$ we have $d(x,x) = 0 < \epsilon$ and so $x$ must be in its own neighbourhood

Keeran Brabazon
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  • Then, if the neighbourhood of x only contains of x itself, is x an interior point of E? Consequently, will E be an open set? – Parul Sep 20 '13 at 17:25
  • Consider that in an open set there are points outside of the set to which a Cauchy sequence can converge. If the set contains only one point there is only one possible Cauchy sequence, which is $x_{i} = x$ for all ${x_{i}}$. Hence $x$ is a closure point of a sequence on the set ${x}$, and so must ${x}$ is closed – Keeran Brabazon Sep 20 '13 at 17:29
  • Basically I want to know how can a finite set E in R2 not be open? Isn't every point of E an interior point of E (since the point will be contained in its own neighbourhood)? – Parul Sep 20 '13 at 17:34
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    You're confusing the subspace topology with the topology of the total space. Each point is open in $E$, yes, but not in $\mathbb{R}^2$. Similarly, a finite set $E \subset \mathbb{R}^2$ is open under its own topology (as is every set with a topology), but is not open in the topology of $\mathbb{R}^2$. – MartianInvader Sep 20 '13 at 17:37
  • That is very confusing indeed. If E is a subset of R2, and any neighbourhood of x is in E, then won't the neighbourhood belong to R2 as well? – Parul Sep 20 '13 at 18:04
  • Being a "neighborhood" depends on what topology you're talking about. A neighborhood in $E$ will be a subset of $\mathbb{R}^2$, but not a neighborhood in $\mathbb{R}^2$, because it leaves out all those in-between points that aren't in $E$ and so it's not open in $\mathbb{R}^2$. A neighborhood of $x$ in $\mathbb{R}^2$ will always contain more than just elements of $E$. – MartianInvader Sep 20 '13 at 18:45