Denumerable partition of a denumerable set where each set in the partition is denumerable. [duplicate]

Suppose that a set $A$ is denumerable. Prove that there is a partition $P$ of $A$ where $P$ is denumerable and every $X \in P$ is also denumerable.
I can see that this can be done but I cannot figure out how to construct it where every $X \in P$ is denumerable and not finite. Any ideas here?
marked as duplicate by Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 19 '15 at 18:08
HINT: Let $A=\{x_n:n\in\Bbb N\}$, and use the inverse of the pairing function.