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 June 3rd, 2018, 10:53 PM #1 Member   Joined: Apr 2017 From: India Posts: 74 Thanks: 0 Real Analysis Do the closed set consists of points other than limit points? If yes, then what are the examples? Last edited by skipjack; June 5th, 2018 at 07:30 AM.
 June 4th, 2018, 01:33 PM #2 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 Isolated points? Thanks from romsek
 June 5th, 2018, 06:58 AM #3 Member   Joined: Apr 2017 From: India Posts: 74 Thanks: 0 Yes points that are apart from limit points in the set. I didn't get what you meant by Isolated points.
 June 5th, 2018, 07:23 AM #4 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic Geometry Every point $x$ of a set is a limit point of that set (even if it's an "isolated point"). Indeed, consider the constant sequence $x, x, \dots$.
June 5th, 2018, 07:51 AM   #5
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 Originally Posted by shashank dwivedi Yes points that are apart from limit points in the set. I didn't get what you meant by Isolated points.
The set $[0,1] \cup \{2\}$ is closed but $2$ is not a limit point. It's an isolated point. It has a neighborhood that contains no other point of the set besides itself.

 June 5th, 2018, 11:14 AM #6 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic Geometry Ah yes, ignore my previous post. I mistakenly took $x$ being a limit point to mean "$x$ is the limit of a sequence of points in the set" (rather than "$x$ is the limit of an eventually non-constant sequence of points in the set"/"every neighbourhood of $x$ contains a point of the set except $x$). Thanks from Joppy
June 5th, 2018, 11:54 AM   #7
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 Originally Posted by cjem Ah yes, ignore my previous post. I mistakenly took $x$ being a limit point to mean "$x$ is the limit of a sequence of points in the set" (rather than "$x$ is the limit of an eventually non-constant sequence of points in the set"/"every neighbourhood of $x$ contains a point of the set except $x$).
This is a tricky point (no pun) that confuses everyone.

An adherent point, also known as a point of closure, is a point whose every neighborhood contains some point of the set. So $2$ is a point of closure of $[0,1] \cup \{2\}$. But it's not a limit point, which requires that every neighborhood of the point contains some point of the set other than the point in question.

June 5th, 2018, 02:23 PM   #8
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 Originally Posted by Maschke This is a tricky point (no pun) that confuses everyone. An adherent point, also known as a point of closure, is a point whose every neighborhood contains some point of the set. So $2$ is a point of closure of $[0,1] \cup \{2\}$. But it's not a limit point, which requires that every neighborhood of the point contains some point of the set other than the point in question.
Yeah, I'm used to the terms "accumulation point" and "point of closure" for the two concepts. I knew limit point referred to one of the two, but settled on the wrong one without bothering to check!

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