By Charles Swartz

ISBN-10: 0824786432

ISBN-13: 9780824786434

In response to an introductory, graduate-level direction given through Swartz at New Mexico nation U., this textbook, written for college students with a reasonable wisdom of aspect set topology and integration idea, explains the foundations and theories of practical research and their purposes, displaying the interpla

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Let {tk) Eel with tk > 0 `dk. 18) by II (sk)II = I I tksk I (11 11 clearly depends on { tk} ). k=1 Consider the sequence {ek} in m0. :jEPI}II = I Itn,I} j=k+1 J so {ek} is I I II-X convergent to 0. However, no subseries of converge to an element of m0 with respect to II II II II k ek can since convergence in clearly implies coordinatewise convergence. 1-space. 20). However, we show in Proposition 10 below that any quasi-normed . if-space is a X-space. We will see later that 4-spaces are useful in obtaining uniform boundedness principles (§9).

So L is not bounded. Recall that a subset metrically bounded if S of a metric space (X, d) is said to be 3 a e X, r> 0 such that S c S(a, r), where S(a, r) = (x : d(a, x) < r) is the sphere with center at a and radius r. Proposition 5. In a quasi-normed space a bounded subset is metrically bounded, but the converse does not hold in general. Proof: Let B be bounded. Then 3 n e IN such that B c nS(0, 1). But nS(0, 1) c S(0, n) so B c S(0, n) is metrically bounded. 21 is metrically bounded, but not bounded by Corollary 4.

Chapter 4 39 Corollary 4. A linear subspace L of a TVS is bounded a L c (0). In particular, no non-trivial subspace of a Hausdorff TVS is bounded. Proof: s: Every neighborhood of 0 contains (0) so (0) is bounded. Suppose 3 a E L\{0}. Then na e L t/n. But (n)na = a f+ 0 since a f (0). So L is not bounded. Recall that a subset metrically bounded if S of a metric space (X, d) is said to be 3 a e X, r> 0 such that S c S(a, r), where S(a, r) = (x : d(a, x) < r) is the sphere with center at a and radius r.