Is the following set connected or disconnected?

Question

Which of the following sets is connected?

1. Space $X = \mathcal{C}[0,1]$ with its usual "sup-norm" topology $$S = \left\{f\in X \mid \int_0^1 f(t) \: dt \neq 0\right\}$$

2. The set $$K = \{f\in\mathcal{C}[0,1] \mid \int_0^{\frac{1}{2}} f(t) \: dt - \int_{\frac{1}{2}}^1 f(t) \: dt = 1\}$$

From my point of view, both are connected and path connected.

For the path connectedness, consider two functions $f$ and $g$ in $S$ or $K$. Let us take $K$; that is, $f$ and $g$ are in $K$. For $t\in[0,1]$, consider $\phi(t): x \mapsto (1-t)f(t) + tg(t)$. The map $t \mapsto \phi(t)$ is continuous path from $f$ to $g$ which bears entirely in $K$, and for all $t\in [0,1]$: $$\int_{0}^{1}\phi(u)\,\mathrm du = (1-t)\int_{0}^{1/2}f(u)\,\mathrm du + t\int_{1/2}^{1}g(u)\,\mathrm du = 0 \ldots$$

Hence both sets $S$ and $K$ satisfy the path convexity, and so both are connected.

Is my answer is correct or not?

Answer 1

$S$ is not path connected ! Let $f(x)=1$ and $g(x)=-2x$. Then $f,g \in S$, but

$$\int_0^1(\frac{1}{2}f(x)+\frac{1}{2}g(x)) dx =0.$$.

$K$ is path connected, but your proof is not correct. Let $f,g \in K$ , $t \in [0,1]$ and let $h:=(1-t)f+tg$.

Then show carefully that $h \in K$.

Answer 2

Your argument proves that $K$ is connected (and, indeed, path-connected). But $S$ is not connected, because you can write $S$ as the disjoint union of $S^+$ and $S^-$, where$$S^+=\left\{f\in\mathcal{C}\bigl([0,1]\bigr)\,\middle|\,\int_0^1f(t)\,\mathrm dt>0\right\}\text{ and }S^-=\left\{f\in\mathcal{C}\bigl([0,1]\bigr)\,\middle|\,\int_0^1f(t)\,\mathrm dt<0\right\}.$$This is a disjoint union and $S^+$ and $S^-$ are open and non-empty. Therefore, $S$ is not connected.

Your argument is wrong because your path contains some function whose integral is $0$, and therefor it does not belong to $S$.