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Let $X$ be all real-valued bounded functions on $[0,1]$ with the supremumsmetric.

Prove this:

a) If it has a countable basis $\{B_i\}$ then $ \{ U_{1/2} (x_i ) = \{x \in X | d(x,x_i ) <1/2 \} \} $ (with a fixed $x_i$ in a $B_i$ for each $i$ ) is an open cover of $X$.

b )We define,for each $a$ in $[0,1]$ $f_a (x) := 1$ , if $a=x$ and $f_a (x)=0$ else. Then $d(f_a , f_b) =0$ if $a=b$ and 1 else. Now from this I have to show that $X$ can't have a countable basis.

Please help me.Thanks

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You wrote: "$f_a (x) := 1$ , if $a=x$ and $f_a (x)$ else." Did you mean "$f_a(x):=0$ else"? –  Martin Sleziak Jan 9 '12 at 13:12
    
As it is your first post here, I think it might be good to mention the policy of this site on homework questions. If this is a homework, you should use homework tag and you should also read this. –  Martin Sleziak Jan 9 '12 at 13:14
    
ok,thank you.i edit. –  Peter Lang Jan 9 '12 at 19:29
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1 Answer

Hint for part a): Given $x\in X$, consider the open ball centered at $x$ of radius $1/2$.

Hint for part b): Every $f_a$ is contained in some $U_{1/2}(x_{i(a)})$. Show that if $a\ne b$, then $U_{1/2}(x_{i(a)})\ne U_{1/2}(x_{i(b)})$. Now note that there are uncountably many distinct $f_a$. What does this imply about the cardinality of $\{ U_{1/2}(x_i) \}$, and thus about the cardinality of the basis?


For part a): Let $x$ be an element in $X$. Consider the open ball $B$ of radius $1/2$ centered at $x$. By the definition of basis, there is a basis element $B_i$ with $x\in B_i$ and $B_i\subset B$. But then, since $x_i\in B_i$, it follows that $x_i\in B$; and thus, $d(x,x_i)<1/2$. So $x\in U_{1/2}(x_i)$.

For part b): If $a\ne b$, then as remarked in your post $d(f_a,f_b)=1$. By the triangle inequality, this implies that $f_a$ and $f_b$ cannot both belong to any open ball of radius $1/2$. In particular, $f_a$ and $f_b$ belong to two different $U_{1/2}(x_j)$.

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thank you david mitra. here is my try: a) let x be arbitrary and $ z \in U_{1/5} (x) $.Chose B_i with $ B_i \leq U_{1/5} (x)$. Then we have $ x_i \in U_{1/5} (x)$. So we have $ d(z,x_i) \leq d(z,x ) + d(x_i ,x) \leq 1/5 + 1/5 < 1/2 $. so $ z \in U_{1/2} (x_i ) $ and $U_{1/5} (x) \leq U_{1/2} (x_i )$ is this correct? b)Is it enough to show $ x_a \neq x_b $ ? i tried several things but please give me another hint to show $U_{1/2}(x_{i(a)})\ne U_{1/2}(x_{i(b)})$. thank you –  Peter Lang Jan 9 '12 at 19:29
    
@Peter Lang I'm not sure about the first part (I think it's ok; but you can do part a) a bit more directly). I added to my post to, hopefully, make things a bit more clear. –  David Mitra Jan 9 '12 at 19:52
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