I have been asked the following question.

The closed unit sphere is not a sequentially compact space. We define the closed unit sphere as $ K: = {f in T (a, b) mid | f | _ infty leq 1 } $

While $ T (a, b) $ is the set of all step functions of $ (a, b) subset mathbb {R} $ Y $ | f | _ infty = sup {| f (x) |: x in (a, b) } $

So I need to find a sequence $ (f_n) subset K $ that it has no converging subsequence that has a limit in K, if I'm correct.

I thought about this problem a lot and came up with this solution …

$ f_n (x): = begin {cases} 1 & text {for} quad x = b \ frac {1} {n} & text {for} quad frac {ba} {r} + a leq x < frac {ba} {r-1} + a quad text {con} quad r in {2, …, n } \ 0 & text {for} quad x = a end {cases} $

Explanation: $ (f_n) $ obviously it's a step function in (a, b) and $ | f_n | leq 1 $. But the limit function $ lim_ {n rightarrow infty} f_n $ it is not a passing function since it has infinite pieces.

Because the sequence boundary converges to $ f_ infty $, all the subsequences converge to $ f_ infty $. Therefore, there is no subsequence in K, and therefore K is not a sequentially compact space.

Is it correct or am I completely on the wrong path? Thanks for helpful answers!