Show that both [itex]Y\cup A[/itex] and [itex]Y\cup B[/itex]

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SUMMARY

The discussion focuses on proving that the sets [itex]Y \cup A[/itex] and [itex]Y \cup B[/itex] are connected given that [itex]Y \subset X[/itex] and [itex]A, B[/itex] form a separation for [itex]X - Y[/itex]. The approach involves assuming a separation [itex]C \cup D[/itex] of [itex]Y \cup A[/itex], where [itex]Y[/itex] is included in either [itex]C[/itex] or [itex]D[/itex]. Utilizing Munkres' lemma from 'Topology - A First Course', it is established that [itex]D, C \cup B[/itex] forms a separation of [itex]X[/itex], leading to the conclusion that no separation [itex]C, D[/itex] can exist, thereby confirming the connectivity of both sets.

PREREQUISITES
  • Understanding of connectedness in topology
  • Familiarity with separations in topological spaces
  • Knowledge of limit points and their properties
  • Basic proficiency in reading mathematical proofs and terminology
NEXT STEPS
  • Study the concept of connectedness in topology, focusing on definitions and examples
  • Review Munkres' 'Topology - A First Course', particularly lemma 1.1 on page 47
  • Explore the implications of separations in connected spaces
  • Investigate limit points and their role in topology
USEFUL FOR

Mathematics students, particularly those studying topology, educators teaching advanced mathematics, and anyone interested in the properties of connected spaces and separations.

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Homework Statement



Suppose Y\subset X and X,Y are connected and A,B form separation for X-Y then, Prove that Y\cup A and Y\cup B are connected.




The Attempt at a Solution



I can show easily that at least one of Y\cup A or Y\cup B is connected. but I don't know how to show the required

Any hints?

Here are some facts that I know about connectdness that may be helpful.

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As a way to proceed, we can suppose that There is a separation C \cup D of,say, Y\cup A So it's clear that Y included in either C or D
 
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I'd say that's good start, and you can choose C to be the one containing Y.
Maybe you can now show that D, BUC form a separation of X? Or something like that.
 
Maths Lover said:
As a way to proceed, we can suppose that There is a separation C \cup D of,say, Y\cup A So it's clear that Y included in either C or D
Yes that is how I would start. Let's call C the part that includes Y.
Next we can use the lemma (1.1 on page 47 of my very old edition of 'Topology - a first course' by Munkres) that if two sets form a separation then neither contains any limit points of the other, So C contains no limit points of D, and also B contains no limit points of D because ##D\subset A## and A,B is a separation of X-Y.

But then ##D, C\cup B## is a separation of ##X## because
##\{x\vert x## is a limit point of ##C\cup B\}\subset \{x\vert x## is a limit point of ##C\}\cup\{x\vert x## is a limit point of ##B\}\subset X-D##

Since ##X## is connected we conclude that there can be no separation ##C,D##.
 

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