Open set

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Topological space

In a topological space [ilmath](X,\mathcal{J})[/ilmath] we have:

  • [ilmath]\forall S\in\mathcal{J} [/ilmath] that [ilmath]S[/ilmath] is an open set. [ilmath]\mathcal{J} [/ilmath] is by definition the set of open sets of [ilmath]X[/ilmath]

Metric space

In a metric space [ilmath](X,d)[/ilmath] there are 2 definitions of open set, however it will be shown that they are equivalent. Here [ilmath]U[/ilmath] is some arbitrary subset of [ilmath]X[/ilmath].

I claim that the following definitions are equivalent:

Definition 1

  • A set [ilmath]U\subseteq X[/ilmath] is open in [ilmath](X,d)[/ilmath] (or just [ilmath]X[/ilmath] if the metric is implicit) if [ilmath]U[/ilmath] is a neighbourhood to all of its points[1], that is to say:
    • [ilmath]U\subseteq X[/ilmath] is open if [math]\forall x\in U\exists\delta_x>0[B_{\delta_x}(X)\subseteq U][/math] - (recall that [ilmath]B_r(x)[/ilmath] denotes the open ball of radius [ilmath]r[/ilmath] centred at [ilmath]x[/ilmath]) or
    • For all [ilmath]x[/ilmath] in [ilmath]U[/ilmath] there is an open ball centred at [ilmath]x[/ilmath] entirely contained within [ilmath]U[/ilmath]

Definition 2

  • A set [ilmath]U\subseteq X[/ilmath] is open in [ilmath](X,d)[/ilmath] (or just [ilmath]X[/ilmath] if the metric is implicit) if[2]:

It is easy to see that these definitions are very similar to each other (these are indeed equivalent is claim 1)

Immediate results

It is easily seen that:

  • [ilmath]\emptyset[/ilmath] is open (claim 2)
  • [ilmath]X[/ilmath] itself is open (claim 3)
  • [ilmath]\text{Int}(U)[/ilmath] is open
    (see interior for a proof of this. The claim is the same as [ilmath]\text{Int}(\text{Int}(U))=\text{Int}(U)[/ilmath] as by the first claim we can use either definition of open, so we use [ilmath]U[/ilmath] is open if [ilmath]U=\text{Int}(U)[/ilmath])

Proof of claims

Claim 1: A set, [ilmath]U\subseteq X[/ilmath] is open according to definition [ilmath]1[/ilmath] [ilmath]\iff[/ilmath] [ilmath]U[/ilmath] is open according to definition [ilmath]2[/ilmath]

TODO: Trivial proof, be bothered to do it

Claim 2: [ilmath]\emptyset[/ilmath] is open

TODO: Trivial proof, be bothered to do it

Claim 3: [ilmath]X[/ilmath] itself is open

TODO: Trivial proof, be bothered to do it

See also


  1. Introduction to Topology - Bert Mendelson
  2. Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene

Old page

Here [math](X,d)[/math] denotes a metric space, and [math]B_r(x)[/math] the open ball centred at [math]x[/math] of radius [math]r[/math]

Metric Space definition

"A set [math]U[/math] is open if it is a neighborhood to all of its points"[1] and neighborhood is as you'd expect, "a small area around".


A set [math]N[/math] is a neighborhood to [math]a\in X[/math] if [math]\exists\delta>0:B_\delta(a)\subset N[/math]

That is if we can puff up any open ball about [ilmath]x[/ilmath] that is entirely contained in [ilmath]N[/ilmath]

Topology definition

In a topological space the elements of the topology are defined to be open sets


A subset [ilmath]N[/ilmath] of a Topological space [ilmath](X,\mathcal{J})[/ilmath] is a neighbourhood of [ilmath]p[/ilmath][2] if:

  • [math]\exists U\in\mathcal{J}:p\in U\wedge U\subset N[/math]

See also


  1. Bert Mendelson, Introduction to Topology - definition 6.1, page 52
  2. Introduction to topology - Third Edition - Mendelson