Undergrad Is there a "Nice" proof that R^2 is not disconnected when we remove 2 points

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SUMMARY

The discussion centers on proving that the space X = R² - {p, q} remains connected after removing two points p and q. Participants emphasize that path-connectedness implies connectedness, which is a standard result at the undergraduate level. A proof is suggested using the contradiction approach, where if X were not connected, it would lead to a contradiction involving the continuous mapping of an interval. The conversation also touches on the challenges of proving the connectedness of the unit interval and the implications of connected sets in topology.

PREREQUISITES
  • Understanding of path-connectedness and connectedness in topology
  • Familiarity with continuous functions and their properties
  • Knowledge of the unit interval and its connectedness
  • Basic concepts of open and closed sets in metric spaces
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  • Study the proof that path-connectedness implies connectedness in topology
  • Learn about the properties of continuous functions and their implications for connected sets
  • Explore the concept of star-shaped sets and their role in proving connectedness
  • Investigate the theorem that products of connected sets are connected and its proof
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Mathematics students, particularly those studying topology, educators teaching undergraduate courses, and anyone interested in the foundational concepts of connectedness in metric spaces.

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TL;DR
Trying to show R^2 -{p,q} is connected.
Are there "nice" ( without heavy machinery) proofs that ## X:=R^2 - \{p,q\} ## is connected? All I can think is using that path-connectedness implies connectedness. So we consider x,y in X and show there is a path joining them. I am looking for an argument at undergrad level, so that I would not have to prove , as in here, that path-connectedness implies connectedness.
 
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Is path connected -> connected really that hard to prove? I would have thought it's a pretty standard undergraduate result.
 
Office_Shredder said:
Is path connected -> connected really that hard to prove? I would have thought it's a pretty standard undergraduate result.
Guess you're right. Maybe I am being too lazy.
 
If X is not connected, and U,V are two disjoint non empty open subsets that cover X, let p be in U and q in V. Then continuously map an interval f:I-->X so that f(0) = p and f(1) = q. Then f^-1(U) and f^-1(V) are disjoint non empty open sets covering I, a contradiction since I is connected.

The point is that I is connected and the image of a connected set is connected, hence any path connected set is connected.

Just out of curiosity, how would you prove to this class that the plane itself is connected without this result?

thinking again about this topic, the hard part of course is to prove the unit interval is connected in the sense of open sets. then after that, proving other sets are connected seems to proiceed naturally via the route outlined here, i.e. iamges of connected sets are conncted and then path connected implies connected. the theorem that products of connected sets are connected seems more difficult, maybe easier to prove products preserve path connectedness.

If you knew star shaped sets are connected, then one could write the twice punctured plane as a union of two such with overlap, but how to prove that?

in gheneral the easiest way to prove facts about sets you know are connected is to use the fact that a set is connected iff every continuous map from it to a 2 point set is constant, but that does not easily help greatly to establish connectivity of a specific set like an interval. or rather, evben with that aid, you still need to use the lub property.
 
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