Basic Topology Proof: y in E Closure if E is Closed

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The discussion focuses on proving that if y is the supremum of a nonempty, bounded above set E of real numbers, then y is in the closure of E if E is closed. The closure of E is defined as the union of E and its limit points. The argument presented states that since y is either in E or its limit points, if E is closed, then all limit points are included in E, making the closure equal to E. There is a question raised about whether the proof adequately demonstrates that y, as the supremum, must be in the closure of E. The conversation emphasizes the need for clarity in proving the relationship between the supremum and the closure of the set.
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Homework Statement



Let E be a nonempty set of real numbers which is bounded above. Let y=sup E. Then y \in E closure. Hence y \in E if E is closed.

Homework Equations



E closure = E' \cup E where E' is the set of all limit points of E.

The Attempt at a Solution



By the definition of closure, y is either in E' or E (maybe in both). If E is closed, E' \subset E and we know that the union of a set and its subset is the set itself. Therefore E closure = E.

Is this a valid proof?
 
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Shouldn't you be proving that if y=sup(E), then y ∈ E closure?
 

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