If $C$ is a convex set then clearly its relative interior is contained in the relative interior of its closure, as it has the same affine hull as its closure. Hence it suffices to show if $x \in$ ri cl $C$ then $x \in$ ri $C$.
By definition $x \in$ ri cl $C$ means there exists $\varepsilon > 0$ such that every point $y \in B(x;\varepsilon) \cap$ aff $C$ is a limit point of $C$ which I guess means $C$ is dense in $B(x;\varepsilon) \cap$ aff $C$? I think from this you can construct line segments in the ball and show that $x \in C$, but I'm still not sure how to show that it is in the relative interior.
According to the book I'm reading and a pdf I found online there should be an easier proof that resembles the proof (in that they both use the line segment principle) that cl ri $C = $ cl $C$. The proof goes as follows:
It suffices to show the cl $C \subset$ cl ri $C$ so suppose $x$ is in cl $C$ and $y \in \text{ri} C$. The line segment principle tells us all the points along the line segment between $x$ and $y$ are in the relative interior of $C$, making $x$ a limit point of the relative interior as desired.
Try as I might, I can't seem to use a similar proof to get the result I want. Any ideas?
Best Answer
Suppose $x \in \text{ri cl } C$. Then by definition there exists an $\varepsilon > 0$ ball around $x$ whose intersection with the affine hull is contained in cl $C$.
The relative interior of $C$ is non-empty so let $y \in \text{ri } C$, $\delta:= \frac \varepsilon {\| x - y \|}$, and define
$$ z:= \Big( 1 + \delta \Big) x - \delta y$$
Clearly $z$ is an affine combination of points in $\text{cl } C$, and it is easy to verify it is within $\varepsilon$ distance of $x$, so we may conclude $z \in \text{cl } C$.
Finally we have
$$ x = \frac 1 {1+\delta} z + \frac \delta {1 + \delta} y$$
i.e. $x$ is a convex combination of $z$ and $y$. Since $z$ is in the closure and $y$ is in the relative interior, by the line segment principle $x$ is in the relative interior, as desired.