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Science Forum Index » Mathematics Forum » path connected or connected?
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| boyandshark |
 Posted: Sun Feb 27, 2005 6:33 pm |
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Hi -
I have a bounded, path connected subset of R^n, let us call it S. Now I
form the closure of S_bar. I am wondering if S_bar (which differs from
S only by the addition of accumulation points of S not contained in S)
can also be shown to be path-connected or connected? The
counter-example I tried coming up with is the graph of sin(1/x) for 0 <
x <=1. Let this graph be our S. It seems (0, 0) is an accumulation
point of this set S, and thus a point in the closure S_bar, but it does
not seem that it can be path connected to any points on the graph S?
Can S_bar nevertheless be shown to be connected?
Thanks
Amit |
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| Justin Young |
| Posted: Sun Feb 27, 2005 6:55 pm |
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"boyandshark" <amitgandhi@gmail.com> wrote in message
news:1109547228.797147.96050@l41g2000cwc.googlegroups.com...
Quote: Hi -
I have a bounded, path connected subset of R^n, let us call it S. Now I
form the closure of S_bar. I am wondering if S_bar (which differs from
S only by the addition of accumulation points of S not contained in S)
can also be shown to be path-connected or connected? The
counter-example I tried coming up with is the graph of sin(1/x) for 0
x <=1. Let this graph be our S. It seems (0, 0) is an accumulation
point of this set S, and thus a point in the closure S_bar, but it does
not seem that it can be path connected to any points on the graph S?
Can S_bar nevertheless be shown to be connected?
Thanks
Amit
It is true in any topological space that if S is connected then so is
closure(S).
Try to prove it! |
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| Jannick Asmus |
| Posted: Mon Feb 28, 2005 1:11 am |
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On 28.02.2005 00:33, boyandshark wrote:
Quote: It seems (0, 0) is an accumulation
point of this set S, and thus a point in the closure S_bar, but it does
not seem that it can be path connected to any points on the graph S?
You are right. To have the closure of S, you need to take all of the
interval {0}x[-1,1] to S. S_bar is definitely not path connected.
J. |
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