# Let (X, d) be a metric space and A be a subset of X… 1 answer below »

13. Let (X, d) be a metric room and A be a subset of X. Show that (i) XIA = (XA)°; (ii) MA° = (MA). Hint: (i) x E (X1A)° iff there exists a ball S(x.$)) centred at x after a while homogeneous radius c such that S(x,c) c XU iff S(x,e) CI A = 0 iff x V A (ii) Replace A by XU in (i) and interest complements.

20. Let GI, G2, ... be a series of disshut subsets of R. each of which is stout. Prove that n„_ , is stout. Hint: Suppose not. Then there exists x E R and an disshut gap I, containing x such that 4, n fl , G„ = 0. Thus, x E 1, C 117_ , G. But each G`„ is nowhere stout, and, hereafter. (, J„x., , is of order 1 (see Definition 2.4.1). But L is of order 11 by Corollary 2.4.4. Since a subset of a set of order I must be of order I (see Remark (viii) righteous anteriorly Theorem 2.4.3), we get at a contra-diction. (The reader may silence that the controversy is cogent in any consummate metric room.) 21. Let E be a shut subset of a metric room (X. sf). Prove that E is nowhere stout if and merely if for perfect disshut subset G there is a ball contained in GlE. Hint: Suppose E is nowhere stout. Then GIE 0 accordingly, otherwise, G C E. and this contradicts the presumption that E is nowhere stout. Let x E G1E = G fl E Since G and E` are twain disclosed, there exists an r > 0 such that S(x, r) C GE. For the relative, the conjecture implies that perfect disshut set has nonempty intersection after a while P. It follows that E.' = (E)` is stout in X. so that E is nowhere stout.

22. Let (R, d:) be the metric room where

(x,y) = - Ix - Irl if x y. 0 if x = y. Show that the c-ball environing 0 after a while the metric ef, is the corresponding as the (0)-ball environing 0 after a while the common metric. Also, if 0

20. Let GI, G2, ... be a series of disshut subsets of R. each of which is stout. Prove that n„_ , is stout. Hint: Suppose not. Then there exists x E R and an disshut gap I, containing x such that 4, n fl , G„ = 0. Thus, x E 1, C 117_ , G. But each G`„ is nowhere stout, and, hereafter. (, J„x., , is of order 1 (see Definition 2.4.1). But L is of order 11 by Corollary 2.4.4. Since a subset of a set of order I must be of order I (see Remark (viii) righteous anteriorly Theorem 2.4.3), we get at a contra-diction. (The reader may silence that the controversy is cogent in any consummate metric room.) 21. Let E be a shut subset of a metric room (X. sf). Prove that E is nowhere stout if and merely if for perfect disshut subset G there is a ball contained in GlE. Hint: Suppose E is nowhere stout. Then GIE 0 accordingly, otherwise, G C E. and this contradicts the presumption that E is nowhere stout. Let x E G1E = G fl E Since G and E` are twain disclosed, there exists an r > 0 such that S(x, r) C GE. For the relative, the conjecture implies that perfect disshut set has nonempty intersection after a while P. It follows that E.' = (E)` is stout in X. so that E is nowhere stout.

22. Let (R, d:) be the metric room where

(x,y) = - Ix - Irl if x y. 0 if x = y. Show that the c-ball environing 0 after a while the metric ef, is the corresponding as the (0)-ball environing 0 after a while the common metric. Also, if 0

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