R^n\{x} is connected

  • Thread starter R.P.F.
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  • #1
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Homework Statement



I am trying to show that R^n\{x} where x could be any point in R^n is connected for n>1. I have been thinking about this for a while but haven't had any luck. Seems like that I'm missing something simple. I tried to construct a continuous map from R^n to R^n\{x}. All I came up was to send x to some other point y, but that map did not seem to be continuous. Any help is appreciated!

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The Attempt at a Solution

 

Answers and Replies

  • #2
CompuChip
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Isn't it easier to show that it is path-connected, which implies connected? You can easily find a path between any two points (take a straight line, if it crosses x then make a wobble around it) - I'll leave it up to you to formalise that argument.

If you insist on using the definition, maybe a proof by contradiction would work. I'm thinking along the lines of "let A and B be open disjoint sets in X = Rn \ {x} whose union is X. Let A be the one containing an open neighbourhood of x, and consider A' = (A [itex]\cup[/itex] \{ x \}) and B. Then A' and B are open sets in Rn and their union is Rn, proving that Rn is disconnected.
 
  • #3
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Isn't it easier to show that it is path-connected, which implies connected? You can easily find a path between any two points (take a straight line, if it crosses x then make a wobble around it) - I'll leave it up to you to formalise that argument.

If you insist on using the definition, maybe a proof by contradiction would work. I'm thinking along the lines of "let A and B be open disjoint sets in X = Rn \ {x} whose union is X. Let A be the one containing an open neighbourhood of x, and consider A' = (A [itex]\cup[/itex] \{ x \}) and B. Then A' and B are open sets in Rn and their union is Rn, proving that Rn is disconnected.
Hey,

Thanks for your ideas! Using the second idea you mentioned, I was able to write up a proof using the definition. The proof was not as simple I expected. So yeah, now I'm going to do the path-connected business. :rofl:
 

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