1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is R with the cofinite topology path connected?

  1. Jul 20, 2008 #1
    u is the usual topology, cf is the cofinite topology.

    yes

    proof:
    pick a and b in (R,cf)

    ((0,1),u) ~ ((a,b),u). then the identity on (a,b) is continuous is because (R,cf) [itex] \subset [/itex] (R,u). map 0->a and 1->b. the fn is continuous at the end points because no subset of the image is open in (R,cf). therefore their exists a path between any a and b in (R,cf).

    is this a valid proof?
     
  2. jcsd
  3. Jul 20, 2008 #2
    I didn't quite get that bit - the image of the map is [a,b] with cofinite topology, which has many open sets - whether or not they're open in the ambient space doesn't affect the continuity of the map in question?

    A clearer approach could be as follows. The identity [tex]\iota:(R,u)\rightarrow(R,cf)[/tex] is continuous. Any path from a to b in the usual topology is given by a continuous function [tex]f:I\rightarrow[a,b][/tex]. Thus the composition [tex]\iota\circ f:(I,u)\rightarrow([a,b],cf)[/tex] is continuous.
     
  4. Jul 21, 2008 #3
    whats the definition of continuous fn? my book has: a fn is continuous from (X,a) ->(Y,b) iff every b open image has an a open preimage in X. hence since no subset of [a,b] is cf open all functions are into [a,b] are continuous.

    i actually think that i was being redundant. i think any function s.t. f(0)=a and f(1)=b is a path in (R,cf) .

    yea yours is good too.
     
  5. Jul 21, 2008 #4
    You have to be careful when dealing with functions onto a subset of the second space.
    For instance, take the set [tex]X=R-\{0\}[/tex] with the topology [tex]\tau=\{\emptyset,R^+,R^-,X\}[/tex]. Then using your definition the interval [tex][a,b]\subseteq(X,\tau)[/tex] has no open sets, so any function [tex]f:(I,u)\rightarrow[a,b][/tex] is continuous. Hence [tex](X,\tau)[/tex] is path connected, which implies it is connected. But the disjoint open sets [tex]R^+,R^-[/tex] cover X, a contradiction.

    The problem here is that a function [tex]f:(X,\tau)\rightarrow(Y,\mu)[/tex] is continuous if every open set in [tex]f(X)[/tex] with topology inherited from Y has open preimage.

    As another example, lets work with the usual topology on R. Consider the function function [tex]f:R\rightarrow R[/tex] which maps x to the first digit of its decimal expansion (e.g. 1.2->2, 0.5->5 etc. ) Then in fact this is a mapping [tex]f:R\rightarrow N[/tex]. But the natural numbers contain no open sets of R, so your definition seems to suggest f is continuous. Of course it's not, at each integer there is a discontinuity (draw the graph). Can you prove f is not continuous using the definition I just gave?
     
  6. Jul 22, 2008 #5
    yea you're right thanks for clearing that up for me. good counter examples btw.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Is R with the cofinite topology path connected?
  1. R^2 or R^3? (Replies: 3)

Loading...