
By Sloughter D.
Read or Download A primer of real analysis PDF
Similar analysis books
Read e-book online Neoliberalismus: Analysen und Alternativen PDF
Der Neoliberalismus hat in den letzten Jahren weite Bereiche unserer Gesellschaft geprägt. Es ist ihm gelungen, zumindest in einem Großteil der medialen Öffe- lichkeit die Legitimität des grundgesetzlich geschützten Sozialstaates – erstmals nach 1945 – zu erschüttern und dessen Säulen ins Wanken zu bringen.
- Success Secrets of Sherlock Holmes: Life Lessons from the Master Detective
- Elliptic integrals
- Advances in Depth Image Analysis and Applications: International Workshop, WDIA 2012, Tsukuba, Japan, November 11, 2012, Revised Selected and Invited Papers
- Harmonic Analysis
- Fourier Analysis on Number Fields
- Analysis of Risks Associated with Nuclear Submarine Decommissioning, Dismantling and Disposal
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.
by Joseph
4.5