any interval subset of R is open and closed
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Two metrics on the same set are said to be topologically equivalent of they have the same open sets. So if an open subset, U contained in M is open with respect to one metric if and only if it is open with respect to the other metric. Another way to think of this is two objects are topologically equivalent if one object can be continuously deformed to the other. To be more precise, a homemorphism, f, between two topological spaces is a continuous bijective map with a continuos inverse. If such a map exists between two spaces, the are topologically equivalent.
Space
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.
Space
Space