Is $M = \{f \in \mathcal{C}[0,1] \mid f(0)=0, \quad f(1)=1\}$ closed, bounded and compact? Consider the subset $M$ of $(\mathcal{C}[0,1], \|| . \||_\infty$), defined as $$ M = \{f \in \mathcal{C}[0,1] \mid f(0)=0, \quad f(1)=1\}.$$
Is it closed? Bounded? Compact?
I know it is bounded (by the extreme value theorem), but I haven't been able to determine the other two conditions.
 A: Consider linear functionals $$\varphi_0(f)=f(0),\qquad \varphi_1(f)=f(1)$$ Both functionals are bounded as $|\varphi_j(f)|\le \|f\|_\infty,$ $j=0,1.$ Therefore $\ker\varphi_j$ are closed, $j=0,1.$ We have $M=x+\ker\varphi_0\cap \ker\varphi_1.$ Thus $M$ is closed. The subset $M$ is an affine  subspace of $C[0,1],$ hence it is not bounded, and therefore noncompact. More explicitly the sequence $f_n(x)=x+nx(1-x)\in M$ is unbounded because $f_n(1/2)={1\over 2}+{n\over 4}.$
A: The set is actually not bounded. Indeed consider the sequence $\{f_j\}_{j\in\mathbb{Z}^+}$ in $M$ given by
$$f_j(x)=\begin{cases}
2jx, & 0\leq x\leq\frac{1}{2},\\
2(1-j)(x-1)+1, &\frac{1}{2}\leq x\leq 1.
\end{cases}$$
Then, for each $j\in\mathbb{Z}^+$, we have that
$$\lVert f_j\rVert_\infty=\sup_{x\in[0,1]}\lvert f_j(x)\rvert=j,$$
and so
$$\lim_{j\to\infty}\lVert f_j\rVert_\infty=\infty,$$
which shows that $M$ is unbounded.
The reason your argument by the Extreme Value Theorem does not work is because what you show is that each $f\in M$ is bounded, not that $M$ itself it bounded.
Now let's consider whether its closed or not. Let $\{f_j\}_{j\in\mathbb{Z}^+}$ be a sequence in $M$ with limit $f$. We need to show that $f\in M$. Indeed since convergence in the uniform norm is just uniform convergence, this means that the sequence converges uniformly to $f$, and so, as each $f_j$ is continuous, it follows by the Uniform Limit Theorem that $f$ is also continuous. Furthermore,
$$f(0)=\lim_{j\to\infty}f_j(0)=0, \quad f(1)=\lim_{j\to\infty}f_j(1)=1.$$
It follows that $f\in M$, and so $M$ is closed.
Finally, as each compact set in a metric space is bounded, and $M$ is not bounded, it follows that $M$ is not compact.
