I know if the set A , the closure of A is connected sure A also connected but the converse Iam not sure
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The comb space C=([0,1]X0) union (KX([0,1]) union (0X{0X1]} where K is the set 1/n where n is an integer. It is made up of vertical lines that make it look like a comb. Each of these vertical lines is joined at the bottom to the y axis. You can see immediately the C is connected since each vertical segment is connected and each vertical segment meets the horizontal segment which is also clearly connected. Now, we need to show it is NOT locally connected. Note the following are equivalent: (TFAE) 1. A space X is locally connected 2. Components of open subsets in X are open ( in X) 3. X has a basis consisting of connected subsets Let V be an open ball with the usual metric in the comb space, which I will call C. Let's put V at the point (0,1/2) and the ball has radius 1/4. The vertical segments of the comb will be the components of V. All of these are open except for ones along the y axis. So we have the {0,y| which is an element of R2 1/4<y<3/4} is not open. This violates condition 2 and we have C is not locally connected. Note the comb space is path connected as is the deleted comb space. But the comb space is not path connected.
Somewhere dense is defined to be the following:Let B, t be a topological space and C ⊂ B. C is somewhere dense if (Cl C)o ≠Ø, the empty set. That is, if the closure of the interior of C has at least one non-empty set.See related links for more information.
They are connected by.......~ Having the word sphere in it.....~ All of them play a big role in our worlds environment~
Which of the following postulates states that a quantity must be equal to itself
A Connected Pyramids have 10 Faces, 12 Vertices, 20 Edges.