Topologically and rationally slice knots

Abstract A knot in $$S^3$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:msup> <mml:mi>S</mml:mi> <mml:mn>3</mml:mn> </mml:msup> </mml:math> is topologically slice if it bounds a locally flat disk in $$B^4$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:msup> <mml:mi>B</mml:mi> <mml:mn>4</mml:mn> </mml:msup> </mml:math> . A knot in $$S^3$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:msup> <mml:mi>S</mml:mi> <mml:mn>3</mml:mn> </mml:msup> </mml:math> is rationally slice if it bounds a smooth disk in a rational homology ball. We prove that the smooth concordance group of topologically and rationally slice knots admits a $$\mathbb {Z}^\infty $$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:msup> <mml:mrow> <mml:mi>Z</mml:mi> </mml:mrow> <mml:mi>∞</mml:mi> </mml:msup> </mml:math> subgroup. All previously known examples of knots that are both topologically and rationally slice were of order two. As a direct consequence, it follows that there are infinitely many topologically slice knots that are strongly rationally slice but not slice.