problem about connectedness Are the following two sets connected?
$(a)$ An infinite set $X$ with topology $\tau$ given by $\tau = \{X,\phi \} \cup \{A \subset X \mid X\setminus A \text{ is a finite set} \}$.
$(b)$ The set $K=\bigg\{f\in C[0,1] \ \bigg| \ \int_0^{1/2} f(t)\,dt -\int_{1/2}^1 f(t)\,dt = 1\bigg\}$.
All I know about connected sets is that if the set cannot be written as the union of two disjoint open sets, then it is connected, otherwise it is disconnected. But I think this definition is not enough to prove the connectivity of those sets here, as their formulation is a bit trickier. So I need a help to find the most efficient and short method to determine the connectedness of these kind of sets.
 A: Hints: 


*

*A topological space $X$ is disconnected if we can find two closed sets $A,B\neq \emptyset$, $A\cap B=\emptyset$  such that $X= A\cup B$. What are the closed sets in your topology?

*Your set is convex, i.e. for every two points $f,g\in K$ the line segment joining $f$ and $g$ is completely contained in $K$. 
A: As to a) this is easy with the definition: a set is connected iff it cannot be written as a disjoint union of two non-empty closed sets (or open, as one is the complement of the other, so if both are closed, then both are open and conversely). The closed sets in $\tau$ are exactly $X$ and the finite sets.
If $X = A \cup B$, $A$ and $B$ disjoint closed and non-empty, then neither can be $X$ (as it forces the other to be $\emptyset$), so both are finite and then so would $X$ be. Contradiction. So $X$ is connected.
in b) $K$ is a subset of a linear space, and there often path-connectedness will do the trick: if $f,g \in K$, then so is $tf + (1-t)g$ for $t \in [0,1]$, try to show this using the definition of $K$.
A: In the first case, in the topology you defined, no two open sets are disjoint(say you have two disjoint open sets, then one is the complement of the other so one of the open set is finite, which is not possible) , hence you cannot write the whole set as a disjoint union of two open sets, hence it's connected.
In the $2$nd one, the set is convex, hence path-connected, hence connected.
