Show that there exists no sequence of functions satisfying the following

  • Thread starter poet_3000
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  • #1
poet_3000
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I found this interesting exercise on a topology book I'm reading, but I don't have a clue what to do.

Show that there is no sequence {g_n} of continuous functions from R to R such that the sequence {(g_n)(x)} is bounded iff x is rational (where R = set of real numbers).
 

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  • #2
lavinia
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I found this interesting exercise on a topology book I'm reading, but I don't have a clue what to do.

Show that there is no sequence {g_n} of continuous functions from R to R such that the sequence {(g_n)(x)} is bounded iff x is rational (where R = set of real numbers).

Every real is the limit of a sequence of rational. The value of a continuous function on this sequence of rationals converges to its value on the limit.
 
  • #3
Tinyboss
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Define [tex]B_n=\{x\in\mathbb{R}:|f_k(x)|>n\:\mathrm{ for some }\:k\}[/tex] and note that each [tex]B_n[/tex] is open. Now the set of points where the sequence is unbounded is [tex]\bigcap_{n\ge0}B_n[/tex]. This is a [tex]G_\delta[/tex] set, and the rationals are not a [tex]G_\delta[/tex] set.

http://en.wikipedia.org/wiki/Gδ_set
 
  • #4
ForMyThunder
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Tinyboss,

You are a clever, clever man. How did you think of something like that?
 

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