Difference between revisions of "Interior point (topology)"
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Latest revision as of 09:48, 30 December 2016
Contents
Definition
Metric space
Given a metric space [ilmath](X,d)[/ilmath] and an arbitrary subset [ilmath]U\subseteq X[/ilmath], a point [ilmath]x\in X[/ilmath] is interior to [ilmath]U[/ilmath][1] if:
- [ilmath]\exists\delta>0[B_\delta(x)\subseteq U][/ilmath]
Relation to Neighbourhood
This definition is VERY similar to that of a neighbourhood. In fact that I believe "[ilmath]U[/ilmath] is a neighbourhood of [ilmath]x[/ilmath]" is simply a generalisation of interior point to topological spaces. Note that:
Claim: [ilmath]x[/ilmath] is interior to [ilmath]U[/ilmath] [ilmath]\implies[/ilmath] [ilmath]U[/ilmath] is a neighbourhood of [ilmath]x[/ilmath]
Proof
- As [ilmath]x[/ilmath] is interior to [ilmath]U[/ilmath] we know immediately that:
- [ilmath]\exists\delta>0[B_\delta(x)\subseteq U][/ilmath]
- We also see that [ilmath]x\in B_\delta(x)[/ilmath] (as [ilmath]d(x,x)=0[/ilmath], so for any [ilmath]\delta>0[/ilmath] we still have [ilmath]x\in B_\delta(x)[/ilmath])
- But open balls are open sets
- So we have found an open set entirely contained in [ilmath]U[/ilmath] which also contains [ilmath]x[/ilmath], that is:
- [ilmath]\exists\delta>0[x\in B_\delta(x)\subseteq U][/ilmath]
- Thus [ilmath]U[/ilmath] is a neighbourhood of [ilmath]x[/ilmath]
This completes the proof.
This implication can only go one way as in an arbitrary topological space (which may not have a metric that induces it) there is no notion of open balls (as there's no metric!) thus there can be no notion of interior point.[Note 1]
Topological space
In a topological space [ilmath](X,\mathcal{J})[/ilmath] and given an arbitrary subset of [ilmath]X[/ilmath], [ilmath]U\subseteq X[/ilmath] we can say that a point, [ilmath]x\in X[/ilmath], is an interior point of [ilmath]U[/ilmath][2] if:
- [ilmath]U[/ilmath] is a neighbourhood of [ilmath]x[/ilmath]
- Recall that if [ilmath]U[/ilmath] is a neighbourhood of [ilmath]x[/ilmath] we require [ilmath]\exists\mathcal{O}\in\mathcal{J}[x\in\mathcal{O}\subseteq U][/ilmath]
Relation to Neighbourhood
We can see that in a topological space that neighbourhood to and interior point of are equivalent. This site (like[2]) defines neighbourhood to as containing an open set with the point in it. However some authors (notably Munkres) do not use this definition and use neighbourhood as a synonym for open set. In this case interior point of and neighbourhood to are not equivalent.
I don't like the term interior point as it suggests some notion of being inside, but the point being in the set is not enough for it to be interior! So I am happy with this and stand by the comments on the neighbourhood page
See also
Notes
- ↑ At least not with this definition of interior point, this probably motivates the topological definition of interior point
