Is R with the cofinite topology path connected?

[ice109]

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) $\subset$ (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?

 

[olliemath]

the fn is continuous at the end points because no subset of the image is open in (R,cf)

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 $$\iota:(R,u)\rightarrow(R,cf)$$ is continuous. Any path from a to b in the usual topology is given by a continuous function $$f:I\rightarrow[a,b]$$. Thus the composition $$\iota\circ f:(I,u)\rightarrow([a,b],cf)$$ is continuous.

 

[ice109]

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?

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) .

A clearer approach could be as follows. The identity $$\iota:(R,u)\rightarrow(R,cf)$$ is continuous. Any path from a to b in the usual topology is given by a continuous function $$f:I\rightarrow[a,b]$$. Thus the composition $$\iota\circ f:(I,u)\rightarrow([a,b],cf)$$ is continuous.

yea yours is good too.

 

[olliemath]

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.

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

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

As another example, let's work with the usual topology on R. Consider the function function $$f:R\rightarrow R$$ 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 $$f:R\rightarrow N$$. 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?

 

[ice109]

yea you're right thanks for clearing that up for me. good counter examples btw.