By N. Bourbaki

This is the softcover reprint of the English translation of 1971 (available from Springer on account that 1989) of the 1st four chapters of Bourbaki's *Topologie générale*. It provides all of the fundamentals of the topic, ranging from definitions. vital sessions of topological areas are studied, uniform buildings are brought and utilized to topological teams. genuine numbers are built and their homes validated. half II, comprising the later chapters, Ch. 5-10, is additionally on hand in English in softcover.

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**Additional resources for General Topology: Chapters 1–4**

**Example text**

I, Theorem I) and therefore a e and let II v, be any elementary IIA se~ I• Conversely,Iet h = (h,) e ITA" containing h; for each LeI: V, II V, contains the point (x,) e II AI and the closure of IT AI' contains: point x, e A,; hence I therefore h lies in I I 47 TOPOLOGICAL STRUCTURES COROLLARY. II XI IT, AI A product of non-empty sets is closed in the product space if and only if A, is closed in I If I is finite, a product in X, for each PROPOSITION 8. I X, for each LeI. II A, ,eI is open provided that A, is open e I; but this is no longer so if I is infinite.

0 g are continuous at z, and let V be a neighbourhood of g (z) in X; by definition, there is a finite subset J of I, and for each LeJ an open subset V, of Y, such that V contains the set n I\(U,) and g(z) belongs to this set. EJ IEJ (1. and the hypothesis implies that each of the sets -; (V,)) is a neighbourhood of z in Z; hence -; (V) is also a neighbourhood of z in Z. This completes the proof. 8' is a base of the topology fjj. The general properties of initial structures (Set Theory, Chapter IV, § 2, no.

In order that the product mapping f: (x,) -+ (];(x,) of n XI leI into II Y lEI I should be continuous at a point a = (a,) it is necessary and sufficient that continuous at al for each LeI. 44 J.. IS PRODUCT SPACE, § 4· r / can be written as x -7 (};(prl(x», so that by Proposition I the condition is sufficient. Conversely, for each x e I let g .... ; then fx = prx f L 0 0 g x is continuous at the point ax, by Proposition I. Since gx it follows that if f is continuous at a, then fx is contin- uous at ax.