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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.
     
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