Proving the existence of a non-monotone continuous function defined on [0,1]Let ( I n ) n ∈ N be the sequence of intervals of [0,1] with rational endpoints, and for every n ∈ N let E n = { f ∈ C [ 0 , 1 ] : f is monotone in I n }. Prove that for every n ∈ N , E n is closed and nowhere dense in ( C [ 0 , 1 ] , d ∞ ). Deduce that there are continuous functions in the interval [0,1] which aren't monotone in any subinterval.For a given n, E n can be expressed as E n = E n ↗ ∪ E n ↙ where E n ↗ and E n ↙ are the sets of monotonically increasing functions and monotonically decreasing functions in E n respectively. I am having problems trying to prove that these two sets are closed. I mean, take f ∈ ( C [ 0 , 1 ] , d ∞ ) such that there is { f k } k ∈ N ⊂ E n ↗ with f k → f. How can I prove f ∈ E n ↗ ?. Suppose I could prove this, then I have to show that E n ¯ ∘ = E n ∘ = ∅. This means that for every f ∈ E n and every r &gt; 0, there is g ∈ B ( f , r ) such that g is not monotone. Again, I am stuck. If I could solve this two points, it's not difficult to check the hypothesis and apply the Baire category theorem to prove the last statement.

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Proving the existence of a non-monotone continuous function defined on [0,1]Let   (      I    n        )          n      ∈              N             be the sequence of intervals of [0,1] with rational endpoints, and for every   n  ∈      N      let       E    n    =  {  f  ∈  C  [  0  ,  1  ]  :  f    is monotone in        I    n    }. Prove that for every   n  ∈      N  ,       E    n   is closed and nowhere dense in   (  C  [  0  ,  1  ]  ,      d    ∞    ). Deduce that there are continuous functions in the interval [0,1] which aren&#039;t monotone in any subinterval.For a given n,       E    n   can be expressed as       E    n    =      E          n      ↗        ∪      E          n      ↙       where       E          n      ↗       and       E          n      ↙       are the sets of monotonically increasing functions and monotonically decreasing functions in       E    n   respectively. I am having problems trying to prove that these two sets are closed. I mean, take   f  ∈  (  C  [  0  ,  1  ]  ,      d    ∞    ) such that there is   {      f    k        }          k      ∈              N              ⊂      E          n      ↗       with       f    k    →  f. How can I prove   f  ∈      E          n      ↗      ?. Suppose I could prove this, then I have to show that                     E        n            ¯        ∘    =      E    n    ∘    =  ∅. This means that for every   f  ∈      E    n   and every   r  &amp;gt;  0, there is   g  ∈  B  (  f  ,  r  ) such that g is not monotone. Again, I am stuck. If I could solve this two points, it&#039;s not difficult to check the hypothesis and apply the Baire category theorem to prove the last statement.

Kali Galloway · Accepted Answer

Step 1It is easy to show that       E    n   is close. Just like what you did, let       E    n    1   be the subset of       E    n   containing all functions increasing on       I    n  . Let   {      f    k    } be a sequence in       E    n   converging to   f  ∈  C  [  0  ,  1  ]. Pick   x  ,  y  ∈      I    n  ,   x  &amp;lt;  y. Then       f    n    (  x  )  ≤      f    n    (  y  ) for all n. Take   n  →  ∞ gives   f  (  x  )  ≤  f  (  y  ). Thus       E    n    1   is closed. Now       E    n   is the union of two close set, so it is close.To show that       E    n   has empty interior, let   f  ∈      E    n  . Assume that f is inceasing (the case when f is decreasing can be done similarly). We perturb f a little bit to get a function h which is not monotone. Let Write       I    n    =  [  a  ,  b  ]. Let   ϵ  &amp;gt;  0 be arbitrary. Let       c    1    ,      c    2    ∈  (  a  ,  b  ) such that       c    1    &amp;lt;      c    2    ,   and      f  (  b  )  −  ϵ  &amp;lt;  f  (      c    2    )  ≤  f  (  b  )  .      c    1    ,      c    2   can be found as f is continuous. Let g be a continuous function on [0,1] with the properties:      |    g      |    ≤  ϵ  ,     g  (  a  )  =  g  (  b  )  =  −  ϵ  ,     g  (      c    1    )  =  g  (      c    2    )  =  0  ,     .Step 2Then   h  =  f  −  g satisfies   d  (  f  ,  h      )    ∞    ≤  ϵ and   h  (  a  )  =  f  (  a  )  −  ϵ  &amp;lt;  f  (  a  )  ≤  f  (      c    1    )  =  h  (      c    1    ) and   h  (      c    2    )  =  f  (      c    2    )  &amp;gt;  f  (  b  )  −  ϵ  =  h  (  b  ). Thus h is not monotone. As f,   ϵ  &amp;gt;  0 are arbitrary,       E    n   contains no open set. Hence       E    n   are nowhere dense. (The conclusion is quite interesting).

Proving the existence of a non-monotone continuous function defined on [0,1]

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