# Construct a Compact set of Real numbers whose limit points form a countable set [duplicate]

K = 0 $\cup$ {1/n : n $\epsilon$ Natural numbers}

It has one limit point which is zero, so it is countable and it is compact. Is this correct?

Also, do you have any ideas of compact sets that come to mind whom you can count there limit points, and they have a countably infinite amount of them?

Let G = 0 $\cup$ {1/n : n $\epsilon$ Natural numbers}

Let {$G_n$} Be a collection of sets Such that N $\cup$ {N + 1/r} For r = 1,2.3.... and N = 0,1,2......

Let me explain, you first start with N = 0. For N = 0 you calculate all the values of r. For N = 0 that will give you a limit point of 0. For N = 1, you do the same, and that should give you a limit point of 1, for N = 2 the same, so on and so on.

I don't know if wrote the proper notation for the set I'm envisioning, but I feel like it gets the job done. What do ya'll think? Also, can you guys think of any other ones?

## marked as duplicate by dustin, Joel Reyes Noche, user147263, Ross Millikan, Jonas Meyer 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(); } ); }); }); Feb 23 '15 at 2:14

• HINT: Start by doing for every integer what you did for $0$. Then think about how to modify that to make it compact. – Brian M. Scott Oct 21 '14 at 20:05
Your answer depends on countable including finite. Some definitions accept it, some do not. An easy way to get countably infinite limit points is $K \cup \frac 1m+\frac1n, m,n \in \Bbb N$ Now each $\frac 1n$ is a limit point.