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Topology: An Inquiry-Based Approach

Steven Schlicker

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Section Open Sets

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Open sets are vitally important in topology. In fact, we will see later that every topological space is completely defined by its open sets. Recall that an open ball is an open set. There are other subsets that every metric space contains, and we might ask if they are open or not.
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We have defined open balls, and open balls are the canonical examples of open sets. In fact, as the following theorem shows, the open balls determine the open sets.
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Proof.

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Let X be a metric space and O a subset of X. To prove this biconditional statement we first assume that O is an open set and demonstrate that O is a union of open balls. Let a∈O. Since O is open, there exists Ο΅a>0 so that B(a,Ο΅a)βŠ†O. We will show that
O=⋃a∈OB(a,Ο΅a).
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By the way we chose Ο΅a, B(a,Ο΅a)βŠ†O for every a∈O. So ⋃a∈OB(a,Ο΅a)βŠ†O. For the reverse containment, let x∈O. Then x∈B(x,Ο΅x) and so xβˆˆβ‹ƒa∈OB(a,Ο΅a). Thus, OβŠ†β‹ƒa∈OB(a,Ο΅a). We conclude that O is a union of open balls if O is an open set.
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The proof of the converse is left for the following activity.
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Activity 8.3.

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Let X be a metric space. To prove the remaining implication of Theorem 8.3, assume that a subset O of X is a union of open balls.
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(b)

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Let x∈O. Why is there an open ball B in O that contains x?
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Theorem 8.3 tells us that every open set is made up of open balls, so the open balls generate all open sets much like a basis of a vector space in linear algebra generates all of the elements of the vector space. For this reason we call the set of open balls in a metric space a basis for the open sets of the metric space. We will discuss this idea in more detail in a subsequent section.
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