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I came across this post, but I wasn't sure if $\{\sin(1/x): x > 0\} \cup \{(0,y): y\in [-1,1]\}$ was equivalent to $\{\sin(1/x): x > 0\} \cup \{(0,0)\}$. By equivalent, I mean is solving the problem approached the same way?

Also, could someone please explain Stefan's answer in the link posted above? It seems much more concise than what I had originally tried.

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What exactly do you mean with equivalence between sets? –  Git Gud Jun 21 '13 at 19:30
    
If you would like have explained a certain answer, then you should leave a comment below that answer. What exactly do want to know? If by equivalence you mean that the connectedness of each space implies connectedness of the other, then that's not exactly true. But it suffices to show that the second space is connected. This implies connectedness of the first one because you just add a connected set which intersects the second space. –  Stefan Hamcke Jun 21 '13 at 20:17
    
The closure of each set is equal to the first set you defined (among other things, the first set is closed), but the sets are not equal. For example, the point $(0,1)$ belongs to the first set you defined but not to the second set you defined. –  Dave L. Renfro Jun 21 '13 at 20:54
    
@DaveL.Renfro: Seeing that the first set is the closure of the second set, is another way to show that connectedness of the second set implies connectedness of the first set. –  Stefan Hamcke Jun 21 '13 at 22:09

1 Answer 1

If you're trying to show that the set you gave is connected, then you could use the following fact (a fun one to prove):

If $A$ is a connected subset of a topological space, and $A\subseteq B\subseteq\overline A,$ then $B$ is connected.

[Show that the closure of a connected set is again connected, and observe that $\overline B=\overline A$ in this circumstance.]

Since $\{(x,\sin(1/x)):x>0\}$ is connected and the topologist's sine curve is its closure, then the fact shows you (simultaneously) that your set and the topologist's sine curve are connected.

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I think that the inclusion should read that $B$ lie between $A$ and its closure. –  jbc Jun 21 '13 at 23:21
    
@jbc: D'oh! Thanks. –  Cameron Buie Jun 21 '13 at 23:23

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