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    so we're working in the topology of pointwise convergence over functions f:[0,1]->real.

    f_n=1-nx for 0\leq x \leq\frac{1}{n}
    0 otherwise.

    wat is the interior, boundary and closure of the set {f_n|neN}

    we think that the interior is the empty set but are unsure about the boundary and closure.
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    srsly.

    (Original post by DFranklin)
    x
    hint please :]
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    What are your definitions of interior, boundary and closure?
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    (Original post by DFranklin)
    What are your definitions of interior, boundary and closure?
    they'd be equivalent to whatever you're thinking of.

    interior of H is the largest open subset of H
    closure of H is the smallest closed superset of H
    boundary of H is the set of points such that any open set containing the point contains a point inside H and a point in T\H
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    I messed that one up on the test. Didn't realise till a few hours later.
    For the closure I think that it's the whole of F[0,1], since a finite (or even countable) intersection of the sub-basis sets  \{ f : a < f(x) < b \} will have some function that goes through every 'window' at x, i.e.  a_i < f(x_i) < b_i , and you can make that function take values  f(x) = 1 - nx when  x \neq x_i
    Hence for every function in F[0,1], any open set including the function will include one that intersects A.

    Then if the interior is empty (I think it is too) the boundary is just  \overline{A} \setminus A^\circ  = \overline{A}.
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    (Original post by benwellsday)
    I messed that one up on the test. Didn't realise till a few hours later.
    For the closure I think that it's the whole of F[0,1], since a finite (or even countable) intersection of the sub-basis sets  \{ f : a < f(x) < b \} will have some function that goes through every 'window' at x, i.e.  a_i < f(x_i) < b_i , and you can make that function take values  f(x) = 1 - nx when  x \neq x_i
    Hence for every function in F[0,1], any open set including the function will include one that intersects A.

    Then if the interior is empty (I think it is too) the boundary is just  \overline{A} \setminus A^\circ  = \overline{A}.
    Good work BWD.
 
 
 

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