MWG state the continuity axiom as follows:

C1. $\succsim$ is continuous if for all $L,L',L''$, the two sets below are both closed: \begin{align} S&=\{\alpha\in[0,1]:\alpha L+(1-\alpha)L''\succsim L'\}\\ T&=\{\alpha\in[0,1]:L'\succsim \alpha L+(1-\alpha)L''\} \end{align}

Other authors (e.g. Kreps, Rubinstein, Levin) use a somewhat different formulation:

C2. $\succsim$ is continuous if for all $L,L',L''$ with $L\succsim L'\succsim L''$, there exists an $\alpha\in[0,1]$ such that \begin{equation} L'\sim \alpha L+(1-\alpha)L'' \end{equation}

Are the two formulations equivalent? It's easy to see how C1 implies C2 (just take $\alpha\in S\cap T$). But I'm not sure how C2 implies C1.

The proofs using C2 usually first establish that, with independence, we have \begin{equation} \beta L+(1-\beta)L'\succ \alpha L+(1-\alpha)L' \end{equation} whenever $L\succ L'$ and $1>\beta>\alpha>0$. Hence, C2 and independence together imply C1. But does C2 imply C1 without independence?


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