By Sloughter D.

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**Additional info for A primer of real analysis**

**Example text**

Note that R is an open set (it is, in fact, the open interval (−∞, +∞)), as is ∅ (it satisfies the definition trivially). 2. Suppose A is a set and, for each α ∈ A, Uα is an open set. 4) α∈A is an open set. Proof. Let x ∈ α∈A Uα . Then x ∈ Uα for some α ∈ A. Since Uα is open, there exists an > 0 such that (x − , x + ) ⊂ Uα . Thus (x − , x + ) ⊂ Uα ⊂ Uα . 5) α∈A Hence α∈A Uα is open. D. 3. 3. Suppose U1 , U2 , . . , Un is a finite collection of open sets. 6) i=1 is open. n Proof. Let x ∈ i=1 Ui .

Proof. Assume C is closed and let U = R \ C. If C = R, then U = ∅, which is open; if C = ∅, then U = R, which is open. So we may assume both C and U are nonempty. Let x ∈ U . Then x is not a limit point of C, so there exists an > 0 such that (x − , x + ) ∩ C = ∅. 12) so U is open. Now suppose U = R \ C is open. If U = R, then C = ∅, which is closed; if U = ∅, then C = R, which is closed. So we may assume both U and C are nonempty. Let x be a limit point of C. Then, for every > 0, (x − , x + ) ∩ C = ∅.

If C = R, then U = ∅, which is open; if C = ∅, then U = R, which is open. So we may assume both C and U are nonempty. Let x ∈ U . Then x is not a limit point of C, so there exists an > 0 such that (x − , x + ) ∩ C = ∅. 12) so U is open. Now suppose U = R \ C is open. If U = R, then C = ∅, which is closed; if U = ∅, then C = R, which is closed. So we may assume both U and C are nonempty. Let x be a limit point of C. Then, for every > 0, (x − , x + ) ∩ C = ∅. 13) > 0 such that (x − , x + ) ⊂ U. 14) Thus x ∈ / U , so x ∈ C and C is closed.

### A primer of real analysis by Sloughter D.

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