**0 contains at least one point less than 0 and therefore not an element of [0,1] to prove it is closed Ex. It will convert the intervals Proof. Problem 4.3: Just apply the de nitions. Furthermore assume that U≠∅ and take any x0∈U. Question: 8) Show That Every Interval Is Path-connected. Proof. I guess the main difficulty is for not closed subsets. Then is connected iff it is not disconnected. Next, we introduce the notion of the neighborhood of a point, which often gives clearer, but equivalent, descriptions of topological concepts than ones that use open intervals. S1 (the unit circle in R2) is connected. Best Answer 100% (1 rating) Previous question Next question Get more help from Chegg. Also, Gerald Edgar's response to the same question says that such sets cannot be totally disconnected, although he does not mention local connectedness. Indeed, if both a∈U and b∈U, then (since U is open) small neighbourhoods of a and b are also contained in U, so (a-ϵ,b+ϵ) is contained in U (for some ϵ>0), but (a,b) was maximal. If X is an interval P is clearly true. Edit I assumed the subspace to be closed. A subset of a line is connected iﬀ it is an interval. Every Interval In R Is Connected. When defining open intervals though, the recoding Then d(x,y)**

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every interval in r is connected 2020