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Because I read different book find there are two definition and they aren't look like a same thing.

Definition 1 from GTM27

A set is open if it's a member of a topology,

and topology is a family of sets which satisfies:

1)the intersection of any two members is still a member.

2)the union of any two members is still a member

Definition 2 from baby rudin

Let X be a metric space. All points and sets mentioned below are understood to be elements and subsets of X.

A set S is open if every point of S is an interior point of S.

and a point p is an interior point of X if there is a neighborhood N of p such that N is contained by X.

A neighborhood of a point p is a set N(P) consisting of all points q such that d(p,q)< r.

definition 2 is complex.

So I try to make it become simpler to find the relation between the two concept.

I thought a set S is open if for every point p of S there is a set N such that N is the subset of X and consist of all points q that d(p,q)

But I still don't find any similarities between the two concept.

In first definition,we know the definition of open set only depends on the concept of set.

But the second definition tell us the definition depends on point and the set.

This is what I am confused now.

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A set S is open if every point of S is an interior point of S means that every point in the set has a neighborhood lying in the set. –  SDevalapurkar Feb 22 at 1:11

2 Answers 2

up vote 2 down vote accepted

The two definitions are slightly different in the sense that they refer to different contexts. Definition 1 is for open sets in a topological space, while definition 2 is for open sets in a metric space. But it turns out that each metric on a space can define a topology on it such, that open sets of the metric space satisfy the properties of definition 1 (for open sets in topological spaces).

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@frabala's answer above is adequate enough according to me but I wish to clarify the following:

"I thought a set S is open if for every point p of S there is a set N such that N is the subset of X and consist of all points q that d(p,q)"

The argument is not entirely wrong. But this is what you should instead gather. A set $S$ is open iff for every point $p \in S$, there is a neighbourhood of $p$ (i.e. a set $N$ - which though should contain an open set in itself for it to be a neighbourhood) entirely contained in $S$ - not $X$. But such a set $N$ is definitely a subset of $X$ too as it is a member of the Topology.

To further cement the distinction between the definitions might want to read this question and its answer.

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