Does there exists a knot $K\subset \mathbb{S}^3$ such that


1. $K$ is **not** slice
2. $\exists W^4$, $\partial W = \mathbb{S}^3$ *rational homology ball*
3. $\exists $ properly embedded smooth disk $(D,\partial D)\to (W,K)$. ?


In other words $K$ is not slice in $B^4$ but is slice in some rational homology ball.



Rational slice knot that is not slice
Yes. The figure-eight knot is an example: it bounds a smooth slice disk in a rational homology ball. This has been proven in a bunch of different ways, going back to the 1980s. Here are a couple of relevant references:


*Fintushel, Ronald; Stern, Ronald J.*, A (\mu)-invariant one homology 3-sphere that bounds an orientable rational ball, Four-manifold theory, Proc. AMS-IMS-SIAM Joint Summer Res. Conf., Durham/N.H. 1982, Contemp. Math. 35, 265-268 (1984). [ZBL0566.57006](https://zbmath.org/?q=an:0566.57006).


*Cha, Jae Choon*, [**The structure of the rational concordance group of knots**](http://dx.doi.org/10.1090/memo/0885), Mem. Am. Math. Soc. 885, 95 p. (2007). [ZBL1130.57034](https://zbmath.org/?q=an:1130.57034).


*Akbulut, Selman; Larson, Kyle*, [**Brieskorn spheres bounding rational balls**](http://dx.doi.org/10.1090/proc/13828), Proc. Am. Math. Soc. 146, No. 4, 1817-1824 (2018). [ZBL1422.57081](https://zbmath.org/?q=an:1422.57081).


However, it remains unknown whether there's a knot is slice in an integer homology ball but not in $B^4$.


