# Every closed set in a separable metric space is the union of a perfect set and a set which is at most countable [duplicate]

Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable.

(Rudin's Principles of Mathematical Analysis, 3rd ed., Chapter 2, Problem 28)

Now I've managed to hit upon a proof using the preceding problem (which can be generalised) as follows:

Let $X$ be a separable metric space, $E \subset X$, $E$ is uncountable, and let $P$ be the set of all condensation points of $E$. Prove that $P$ is perfect and that at most countably many points of $E$ are not in $P$. In other words, show that $P^c \cap E$ is at most countable.

Now what bothers me is what Rudin states as the corollary to the result he asks us to prove in Problem 28. Here's what he states:

Every countable closed set in $\mathbb{R}^k$ has isolated points.

How to prove this from the original result?

Can we also state the following?

Every countable closed set in a separable metric space has isolated points.

## marked as duplicate by dustin, John Gowers, Robert Cardona, Did real-analysis 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(); } ); }); }); Mar 14 '15 at 19:15

• What about the rationals? Always ask yourself about the rationals. – Carl Mummert Nov 10 '14 at 2:49
• @Carl, I'm not able to see your point. So could you please elaborate? – Saaqib Mahmood Nov 11 '14 at 14:17
• @SaaqibMahmuud the rationals are a dense separable set wich is not closed so it doesnt have any isolated points. – Aram Nov 11 '14 at 14:31
• @Saaqib Mahmuud: If we consider just the rationals as their own separable metric space, they are a countable closed set without isolated points. – Carl Mummert Nov 11 '14 at 14:40
• @Aram: see above, please – Carl Mummert Nov 11 '14 at 14:41

Every countable closed set in $\mathbb{R}^k$ has isolated points.
HINT: Negate it (that is there is no isolated point) then surround two different points by a closed ball then pick $2$ different points inside each ball surround again the $4$ by a smaller closed ball (that is none of the points is inside any of the others balls), and repeat, then apply the Nested Intersection Theorem (Remember a closed subset of a complete space is complete) to any sequence of nested balls and deduce you have a bijection of your limit points with $\{0,1\}^{\aleph} \sim\aleph^{\aleph}$
Suppose your set doesnt have any isolated points, then it is a perfect set of $\mathbb{R}^k$ but this would mean that is not countable, contradiction.