Piece-Wise Function

Question

Give an example of a function $f$ whose domain is the closed interval $[0,1]$ such that $f$ is bounded but does not attain its upper bound (i.e. there is no $x_1$ that exists in $[0, 1]$ such that $f(x) \leq f(x_1)$ for all $x$ that exist in $[0, 1]$).

I have no clue, need help.

Here's a theorem for you (Heine-Borel): every continuous function obtains its maximum (and minimum) on a closed and bounded set, e.g. on closed finite intervals. This suggests you should look to discontinuous functions (and motivates David's response). — Cameron Williams, Apr 15 '14 at 23:13

score 0 · Answer 1 · answered Apr 15 '14 at 23:56

I must admit that I don't see how changing a continuous function at a single point on its compact domain would produce such a function.

However, as Cameron noted, we should look among discontinuous functions. On the top of my head, I'd define $f:\left[ {0,1} \right] \to \mathbb{R}$ by

$f\left( x \right) = \left\{ \begin{gathered} x & ,x \in \mathbb{Q} \cap \left[ {0,1} \right] \\ 1 + x & ,x \in \left( {\mathbb{R}\backslash \mathbb{Q}} \right) \cap \left[ {0,1} \right] \\ \end{gathered} \right.$.

$0 \leqslant f\left( x \right) \leqslant 2,\forall x \in \left[ {0,1} \right]$, so it's bounded, $\mathop {\sup }\limits_{x \in \left[ {0,1} \right]} f\left( x \right) = 2$ and I leave to you to show that the supremum is not attained.

Piece-Wise Function

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