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How is it different from an interior point and boundary point combined together? Isn't a limit point just all the interior points of a set and also the boundary points of the set? Or am I missing something from the definition (our class defined it as a point whose neighborhood contains a point in the set that's not the point itelf)? How is it possible that a neighborhood of a point contains only that point in the set and no other points?

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    $\begingroup$ You misinterpreted the statement in your class. The statement means : In every neighborhood of the limit point, there is a point in the set different from the limit point. So, no matter how small the neighborhood is, there are still (infinitely many) points different from the limit point. $\endgroup$ – Peter Feb 6 '15 at 0:13
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    $\begingroup$ The statement does not mean that there is only one point of the set in the neighborhood and no others. It only means that there is AT LEAST one such point. $\endgroup$ – Peter Feb 6 '15 at 0:14
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    $\begingroup$ On your first question, the boundary includes isolated points. A limit point for a set cannot also be an isolated point for that same set. $\endgroup$ – Matt A Pelto Feb 6 '15 at 0:17
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    $\begingroup$ Consider the singleton set $\{0\}$ on the real line. This set has no limit points, but is its own boundary. $\endgroup$ – MartianInvader Feb 6 '15 at 0:18
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    $\begingroup$ "How is it possible that a neighborhood of a point contains only that point in the set and no other points?" -- It could be an isolated point. Consider $(0,1) \cup \{2\}$, the open unit interval unioned with the singleton point 2. In this case 2 is an isolated point. It's a member of the set, but it's not a limit point because it has a neighborhood that does not intersect any other point of the set. $\endgroup$ – user4894 Feb 6 '15 at 0:44
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Collecting comments that are partial answers:

Pelto: On your first question, the boundary includes isolated points. A limit point for a set cannot also be an isolated point for that same set.

user4894: "How is it possible that a neighborhood of a point contains only that point in the set and no other points?" -- It could be an isolated point. Consider $(0,1)∪{2}$, the open unit interval unioned with the singleton point 2. In this case 2 is an isolated point. It's a member of the set, but it's not a limit point because it has a neighborhood that does not intersect any other point of the set.

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