This example is to show that a connected topological space need not be path-connected.(adsbygoogle = window.adsbygoogle || []).push({});

S={ (t,sin(1/t)): 0 <t <= 1 }

A={ (0,t): -1 <= t <= 1 }

let T = S U A

with the topology induced from R^2.

I show T is not path-connected. Assume to the contrary that there exists a path p:[0,1]-->T with p(0)=(1/pi, 0) and p(1)=(0,0). Let

k = Inf{ t in [0,1] : p(t) in A}.

Then p([0,k]) contains at most one point of A. I want to show Closure of p([0,k]) contains all of A. So p([0,k]) is not closed, and therefore not compact. But p is continuous and [0,k] is compact, so p([0,k]) must be compact,which is a contradiction.

How do I show Closure of p([0,k]) contains all of A?

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# Homework Help: Topologist's sine curve is not path connected

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