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While searching for help as to how to prove that a set is open/closed, I stumbled across a stackexchange post from 2011 which covered this topic. The original poster (Vafa Khalighi) asked for a proof to the following:

"Let (X,d) be an arbitrary metric space. Prove that any set which contains a finite number of points {x1,x2,…,xn} is closed."

In response to this, Weltschmerz gave the proof:

"let's first prove that a one-point set is closed. Fix the point x, for the set {x}. Now consider any other point yy. If d(x,y)=ϵ, take B(y,ϵ/2). This open ball does not intersect {x} (why?). This means that the complement of {x} is open, which implies the set is closed. Now you know that any finite union of closed sets is closed (straight from the definition of topology by application of De Morgan's laws), and the result follows."

Which was accepted as the correct answer (and is similar to the proof given in my uni solutions for a similar question)

I unfortunately still don't understand this, however, as surely one could use this same logic to prove that an open ball is closed?

Take the open ball B(x,r) and any other point y outside the ball. If d(x,y) = d, take B(y, d-r). This open ball does not intersect B(x,r). This means that the complement of B(x,r) is open, which implies the set is closed.

Why does this not work when it seems to follow the same logic as the original proof? Clearly it doesn't since the open ball is not closed.

Thanks for any help!

Original post: The general argument to prove a set is closed/open

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Tom Procter
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    What happens if $d(x,y) = r$? Since we are talking about the open ball, $y$ will not be in this set, but $B(y, d-r)$ in that case is just one point, which does not constitute an open ball. Hence, these are (in fact, only those) points outside the open ball, all of whose neighbourhoods will intersect the open ball. – Sarvesh Ravichandran Iyer Mar 31 '17 at 01:37
  • I see! Thank you! And in the original proof there was no way for something like this to occur and so the proof holds? – Tom Procter Mar 31 '17 at 01:46
  • Exactly, there was no way something like this could happen. However, suppose I included the set $d(x,y) =r$ (the sphere) with the open ball of radius $r$, I get the closed ball of radius $r$, and you can check this is closed using the argument "if a point is outside it, I can find a ball around it that is away from the closed set". – Sarvesh Ravichandran Iyer Mar 31 '17 at 01:48
  • Hi again. This is pretty much a perfect first question. Way to go. Welcome to the site. Don't be a stranger. – The Count Mar 31 '17 at 02:34

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The issue is that if $y \notin B(x,r)$ you have $d=d(x,y) \geq r$. In particular, if $y$ is a point for which $d=d(x,y)=r$, your ball $B(x, d-r)$ has a radius of $0$, while the definition of open sets asks for positive radius.

N. S.
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  • Ahhh and we require the ability to draw an open ball around every point not in B(x,r) in order to show that its complement is open? – Tom Procter Mar 31 '17 at 01:44
  • @TomProcter Exactly. Intuitively, your proof works for the points not on the boundary, but fails on the boundary, which if you consider for a moment makes sense. – N. S. Mar 31 '17 at 01:45
  • Perfect! That makes loads of sense, thanks! – Tom Procter Mar 31 '17 at 01:47