Since the RHS is a closed subset of Z containing Y, it must contain the LHS.Ĭonversely, since LHS is closed in Z it must be of the form C ∩ Z for some closed subset C of X. If are subsets of a topological space X, then the closure of Y in Z is the intersection of Z with the closure of Y in X. Next, the closure of Y in a smaller subspace can be deduced from the closure in a bigger space.
For the second inclusion: since for each i, we have Since this holds for each i, we’re done. ♦. For the first inclusion: since for each i, we have Since this holds for every i, we get the desired inclusion. That being said, we do at least have a one-sided inclusions: On the other hand, cl( Y) ∩ cl( Z) = R ∩ R = R. Then Y ∩ Z is empty, and so is cl( Y ∩ Z). let Y = Q and Z = R– Q be subsets of the real line R. if in X= R, then since each is closed, we have which isn’t closed, so it cannot possibly beĪlso, the result is not true for intersection. Since the RHS is closed and contains, it must contain the LHS also.Ĭonversely, since we have Similarly, so the LHS contains the RHS. ♦ Unions and Intersections of Closuresįor any subsets However, this is as far as we go, as we cannot extend this to the union of infinitely many sets. On the other hand, since cl( Y) is closed, This gives which completes the proof. To complete the proof, since C is a closed subset containing Y, it also contains cl( Y). Thus so X– C is a union of open subsets and is thus open. Since U is an open subset of X containing y, we have U ∩ Y containing a point z ≠ y, which contradicts what we just proved. Next we wish to show Suppose on the contrary so in particular y is a point of accumulation of Y. If then in particular x is not a point of accumulation of Y, so there exists an open subset containing x such that U ∩ Y contains at most x. We have where Y acc is the collection of all points of accumulation of Y. The closure of Y can be characterised as follows. Then Y = is already closed in X so cl( Y) = Y, while the closed ball N(0, 2)* = X ≠ cl( Y). For a rather pathological example, consider and take the open ball Y = N(0, 2). 
However, the closure of the open ball is not the closed ball in general. Warning: in a metric space X, the closed ball is a closed subset since the complement (which contains all x such that d( x, a) > ε) is open in X. Take the set of rationals Y = Q in X = R.Take the half-open interval The closure is.Indeed, cl( Z) is a closed subset of X containing Z, and hence Y. This justifies our calling cl( Y) the smallest closed subset containing Y. if then since cl( Y) is the intersection of many sets, including C, we must have.: since an intersection of closed subsets is still closed.The closure satisfies the following properties:
If there’s any possibility of confusion, we denote it by cl X( Y) instead. Note that the notation cl( Y) does not indicate the ambient space X. Let Σ be the collection of all closed subsets containing Y.We define cl( Y) to be the “smallest” closed subset containing Y. Suppose Y is a subset of a topological space X.
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