Thread: What is the definition of a convex subset of a topological space

I'm trying to show that the Least Core of is a convex set whenever is a convex and closed set in a topological vector space.
The definition of the Least Core is:
Where is the maximum operator and M is the index set of all the funtions on


But what is the definition of a convex subset of a topological space that I should use?
Can I just use this notion:
Let be a vector space (over R or C). A subset S of V
is convex if for all points x,y in S, the line segment
is also in S.


Btw this is no homework, so I don't know exactly what definition to use.

Solved it.


Let $S$ be the Least Core of $(\Pi,F)$. Let $x_1,x_2 \in S$. Show for an arbitrary $\alpha \in (0,1) \,\alpha x_1 + (1-\alpha) x_2$ is contained in $S$.
$x_1,x_2\in S$ so $\bigvee_{j\in M}F_j(x_1)\leq\bigvee_{j\in M}F_j (y)\,, \forall y\in \Pi$ and $\bigvee_{j\in M}F_j(x_2)\leq\bigvee_{j\in M}F_j (y)\,, \forall y\in \Pi$
Because the $F_j$'s are convex we know$F_j(\alpha x_1+(1-\alpha )x_2)\leq \alpha F_j(x_1) + (1-\alpha)F_j(x_2)$. Without loss of generality we may assume $\bigvee F_j(x_2)\geq \bigvee F_j(x_1)$, so: $\bigvee F_j(\alpha x_1+(1-\alpha )x_2)\leq \bigvee F_j(x_2)$ and because $x_2$ lies in the least core, so does $\alpha x_1+(1-\alpha )x_2$\\

The definition of the Least Core is:
Where is the maximum operator and M is the index set of all real funtions F on


How to proof that LC is closed? I tried proving it using a sequence in LC but it made no sense