Determining the space complexity of $x$-obstruction-free $k$-set agreement for $xleq k$ is an open problem. In $x$-obstruction-free protocols, processes are required to return in executions where at most $x$ processes take steps. The best known upper bound on the number of registers needed to solve this problem among $n>k$ processes is $n-k+x$ registers. No general lower bound better than $2$ was known. We prove that any $x$-obstruction-free protocol solving $k$-set agreement among $n>k$ processes uses at least $lfloor(n-x)/(k+1-x)rfloor+1$ registers. Our main tool is a simulation that serves as a reduction from the impossibility of deterministic wait-free $k$-set agreement: if a protocol uses fewer registers, then it is possible for $k+1$ processes to simulate the protocol and deterministically solve $k$-set agreement in a wait-free manner, which is impossible. A critical component of the simulation is the ability of simulating processes to revise the past of simulated processes. We introduce a new augmented snapshot object, which facilitates this. We also prove that any space lower bound on the number of registers used by obstruction-free protocols applies to protocols that satisfy nondeterministic solo termination. Hence, our lower bound of $lfloor(n-1)/krfloor+1$ for the obstruction-free ($x=1$) case also holds for randomized wait-free free protocols. In particular, this gives a tight lower bound of exactly $n$ registers for solving obstruction-free and randomized wait-free consensus. Finally, our new techniques can be applied to get a space lower of $lfloor n/2rfloor+1$ for $epsilon$-approximate agreement, for sufficiently small $epsilon$. It requires participating processes to return values within $epsilon$ of each other. The best known upper bounds are $lceillog(1/epsilon)rceil$ and $n$, while no general lower bounds were known.
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