Every bounded set $Esubseteqmathbb{R}^2$ is contained in a disc of minimal radius
$begingroup$
Consider the following proof from the book "The Banach-Tarski Paradox" by Stan Wagon:
In the second paragraph he seems to implicitely use that every bounded set $Esubseteqmathbb{R}^2$ is contained in a unique disc of minimal radius. Now the uniqueness of such a disc is actually easy to see, but I don't see why it must necessarily exist.
real-analysis vector-spaces metric-spaces
asked Jan 3 at 20:51
AlexAlex
894
$endgroup$
add a comment |
$begingroup$
Consider the following proof from the book "The Banach-Tarski Paradox" by Stan Wagon:
In the second paragraph he seems to implicitely use that every bounded set $Esubseteqmathbb{R}^2$ is contained in a unique disc of minimal radius. Now the uniqueness of such a disc is actually easy to see, but I don't see why it must necessarily exist.
real-analysis vector-spaces metric-spaces
asked Jan 3 at 20:51
AlexAlex
894
$endgroup$
$begingroup$
What's your definition of bounded? Is your problem that there will be A disc or that there will be a MINIMAL-radius disc?
$endgroup$
– Gaffney
Jan 3 at 21:02
$begingroup$
The minimal part. In other words why is there a circle which realizes the inf of radii of enclosing circles
$endgroup$
– Alex
Jan 3 at 21:05
add a comment |
$begingroup$
Consider the following proof from the book "The Banach-Tarski Paradox" by Stan Wagon:
In the second paragraph he seems to implicitely use that every bounded set $Esubseteqmathbb{R}^2$ is contained in a unique disc of minimal radius. Now the uniqueness of such a disc is actually easy to see, but I don't see why it must necessarily exist.
real-analysis vector-spaces metric-spaces
asked Jan 3 at 20:51
AlexAlex
894
$endgroup$
Consider the following proof from the book "The Banach-Tarski Paradox" by Stan Wagon:
In the second paragraph he seems to implicitely use that every bounded set $Esubseteqmathbb{R}^2$ is contained in a unique disc of minimal radius. Now the uniqueness of such a disc is actually easy to see, but I don't see why it must necessarily exist.
real-analysis vector-spaces metric-spaces
real-analysis vector-spaces metric-spaces
asked Jan 3 at 20:51
AlexAlex
894
asked Jan 3 at 20:51
AlexAlex
894
asked Jan 3 at 20:51
AlexAlex
894
asked Jan 3 at 20:51
AlexAlex
894
asked Jan 3 at 20:51
AlexAlex
894
894
$begingroup$
What's your definition of bounded? Is your problem that there will be A disc or that there will be a MINIMAL-radius disc?
$endgroup$
– Gaffney
Jan 3 at 21:02
$begingroup$
The minimal part. In other words why is there a circle which realizes the inf of radii of enclosing circles
$endgroup$
– Alex
Jan 3 at 21:05
add a comment |
$begingroup$
What's your definition of bounded? Is your problem that there will be A disc or that there will be a MINIMAL-radius disc?
$endgroup$
– Gaffney
Jan 3 at 21:02
$begingroup$
The minimal part. In other words why is there a circle which realizes the inf of radii of enclosing circles
$endgroup$
– Alex
Jan 3 at 21:05
$begingroup$
What's your definition of bounded? Is your problem that there will be A disc or that there will be a MINIMAL-radius disc?
$endgroup$
– Gaffney
Jan 3 at 21:02
$begingroup$
What's your definition of bounded? Is your problem that there will be A disc or that there will be a MINIMAL-radius disc?
$endgroup$
– Gaffney
Jan 3 at 21:02
$begingroup$
The minimal part. In other words why is there a circle which realizes the inf of radii of enclosing circles
$endgroup$
– Alex
Jan 3 at 21:05
$begingroup$
The minimal part. In other words why is there a circle which realizes the inf of radii of enclosing circles
$endgroup$
– Alex
Jan 3 at 21:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's suppose that $D(p_0,r_0)$ is a closed disc with center $p_0$ and radius $r_0$ that contains $E$.
Let $CR subset mathbb R^2 times [0,infty) subset mathbb R^3$ be the set of ordered pairs $(q,s)$ such that the closed disc $D(q,s)$ contains $E$ and such that $s le r_0$.
Once you believe that the set $CR$ is closed and bounded, it follows that $CR$ is compact, and therefore the projection function $CR mapsto [0,infty)$ has compact image and hence has a minimum.
The set $CR$ is bounded, because if $(p,r) in CR$ then $r le r_0$, and then by picking any $q in E$ one sees that $d(p,p_0) le d(p,q) + d(q,p_0) le r + r_0 le 2 r_0$.
The set $CR$ is closed, because if $(p,r) in mathbb R^2 times [0,infty)$ is a limit of a sequence $(p_i,r_i) in CR$ then for each point $q in E$ we have $d(q,p_i) le r_i$, and since distance is continuous we may pass to the limit to deduce that $d(q,p) le r$.
answered Jan 3 at 21:14
Lee MosherLee Mosher
49k33684
$endgroup$
add a comment |
$begingroup$
First note, that C and its interior is a closed disc.
This is a little bit messy, but: You take the closure. (Call this F.) This must be a subset of the smallest (closed) disc, if one exists. Now the closure is closed and bounded, so it's compact, and there are two points that realize the diameter. (Call these x and y.) If you take the closed disc for which this diameter is the diameter of the disc, I claim it would be such a smallest closed disc.
First if you had a smaller closed disc, then it's subset, F, would have a bigger diameter, a contradiction. Second, if there were points outside of this disc in F, you could use triangle inequality to argue that that point along with either x or y form a longer diameter in F.
answered Jan 3 at 21:17
GaffneyGaffney
1,403712
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add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's suppose that $D(p_0,r_0)$ is a closed disc with center $p_0$ and radius $r_0$ that contains $E$.
Let $CR subset mathbb R^2 times [0,infty) subset mathbb R^3$ be the set of ordered pairs $(q,s)$ such that the closed disc $D(q,s)$ contains $E$ and such that $s le r_0$.
Once you believe that the set $CR$ is closed and bounded, it follows that $CR$ is compact, and therefore the projection function $CR mapsto [0,infty)$ has compact image and hence has a minimum.
The set $CR$ is bounded, because if $(p,r) in CR$ then $r le r_0$, and then by picking any $q in E$ one sees that $d(p,p_0) le d(p,q) + d(q,p_0) le r + r_0 le 2 r_0$.
The set $CR$ is closed, because if $(p,r) in mathbb R^2 times [0,infty)$ is a limit of a sequence $(p_i,r_i) in CR$ then for each point $q in E$ we have $d(q,p_i) le r_i$, and since distance is continuous we may pass to the limit to deduce that $d(q,p) le r$.
answered Jan 3 at 21:14
Lee MosherLee Mosher
49k33684
$endgroup$
add a comment |
$begingroup$
Let's suppose that $D(p_0,r_0)$ is a closed disc with center $p_0$ and radius $r_0$ that contains $E$.
Let $CR subset mathbb R^2 times [0,infty) subset mathbb R^3$ be the set of ordered pairs $(q,s)$ such that the closed disc $D(q,s)$ contains $E$ and such that $s le r_0$.
Once you believe that the set $CR$ is closed and bounded, it follows that $CR$ is compact, and therefore the projection function $CR mapsto [0,infty)$ has compact image and hence has a minimum.
The set $CR$ is bounded, because if $(p,r) in CR$ then $r le r_0$, and then by picking any $q in E$ one sees that $d(p,p_0) le d(p,q) + d(q,p_0) le r + r_0 le 2 r_0$.
The set $CR$ is closed, because if $(p,r) in mathbb R^2 times [0,infty)$ is a limit of a sequence $(p_i,r_i) in CR$ then for each point $q in E$ we have $d(q,p_i) le r_i$, and since distance is continuous we may pass to the limit to deduce that $d(q,p) le r$.
answered Jan 3 at 21:14
Lee MosherLee Mosher
49k33684
$endgroup$
add a comment |
$begingroup$
Let's suppose that $D(p_0,r_0)$ is a closed disc with center $p_0$ and radius $r_0$ that contains $E$.
Let $CR subset mathbb R^2 times [0,infty) subset mathbb R^3$ be the set of ordered pairs $(q,s)$ such that the closed disc $D(q,s)$ contains $E$ and such that $s le r_0$.
Once you believe that the set $CR$ is closed and bounded, it follows that $CR$ is compact, and therefore the projection function $CR mapsto [0,infty)$ has compact image and hence has a minimum.
The set $CR$ is bounded, because if $(p,r) in CR$ then $r le r_0$, and then by picking any $q in E$ one sees that $d(p,p_0) le d(p,q) + d(q,p_0) le r + r_0 le 2 r_0$.
The set $CR$ is closed, because if $(p,r) in mathbb R^2 times [0,infty)$ is a limit of a sequence $(p_i,r_i) in CR$ then for each point $q in E$ we have $d(q,p_i) le r_i$, and since distance is continuous we may pass to the limit to deduce that $d(q,p) le r$.
answered Jan 3 at 21:14
Lee MosherLee Mosher
49k33684
$endgroup$
Let's suppose that $D(p_0,r_0)$ is a closed disc with center $p_0$ and radius $r_0$ that contains $E$.
Let $CR subset mathbb R^2 times [0,infty) subset mathbb R^3$ be the set of ordered pairs $(q,s)$ such that the closed disc $D(q,s)$ contains $E$ and such that $s le r_0$.
Once you believe that the set $CR$ is closed and bounded, it follows that $CR$ is compact, and therefore the projection function $CR mapsto [0,infty)$ has compact image and hence has a minimum.
The set $CR$ is bounded, because if $(p,r) in CR$ then $r le r_0$, and then by picking any $q in E$ one sees that $d(p,p_0) le d(p,q) + d(q,p_0) le r + r_0 le 2 r_0$.
The set $CR$ is closed, because if $(p,r) in mathbb R^2 times [0,infty)$ is a limit of a sequence $(p_i,r_i) in CR$ then for each point $q in E$ we have $d(q,p_i) le r_i$, and since distance is continuous we may pass to the limit to deduce that $d(q,p) le r$.
answered Jan 3 at 21:14
Lee MosherLee Mosher
49k33684
answered Jan 3 at 21:14
Lee MosherLee Mosher
49k33684
answered Jan 3 at 21:14
Lee MosherLee Mosher
49k33684
answered Jan 3 at 21:14
Lee MosherLee Mosher
49k33684
49k33684
add a comment |
add a comment |
$begingroup$
First note, that C and its interior is a closed disc.
This is a little bit messy, but: You take the closure. (Call this F.) This must be a subset of the smallest (closed) disc, if one exists. Now the closure is closed and bounded, so it's compact, and there are two points that realize the diameter. (Call these x and y.) If you take the closed disc for which this diameter is the diameter of the disc, I claim it would be such a smallest closed disc.
First if you had a smaller closed disc, then it's subset, F, would have a bigger diameter, a contradiction. Second, if there were points outside of this disc in F, you could use triangle inequality to argue that that point along with either x or y form a longer diameter in F.
answered Jan 3 at 21:17
GaffneyGaffney
1,403712
$endgroup$
add a comment |
$begingroup$
First note, that C and its interior is a closed disc.
This is a little bit messy, but: You take the closure. (Call this F.) This must be a subset of the smallest (closed) disc, if one exists. Now the closure is closed and bounded, so it's compact, and there are two points that realize the diameter. (Call these x and y.) If you take the closed disc for which this diameter is the diameter of the disc, I claim it would be such a smallest closed disc.
First if you had a smaller closed disc, then it's subset, F, would have a bigger diameter, a contradiction. Second, if there were points outside of this disc in F, you could use triangle inequality to argue that that point along with either x or y form a longer diameter in F.
answered Jan 3 at 21:17
GaffneyGaffney
1,403712
$endgroup$
add a comment |
$begingroup$
First note, that C and its interior is a closed disc.
This is a little bit messy, but: You take the closure. (Call this F.) This must be a subset of the smallest (closed) disc, if one exists. Now the closure is closed and bounded, so it's compact, and there are two points that realize the diameter. (Call these x and y.) If you take the closed disc for which this diameter is the diameter of the disc, I claim it would be such a smallest closed disc.
First if you had a smaller closed disc, then it's subset, F, would have a bigger diameter, a contradiction. Second, if there were points outside of this disc in F, you could use triangle inequality to argue that that point along with either x or y form a longer diameter in F.
answered Jan 3 at 21:17
GaffneyGaffney
1,403712
$endgroup$
First note, that C and its interior is a closed disc.
This is a little bit messy, but: You take the closure. (Call this F.) This must be a subset of the smallest (closed) disc, if one exists. Now the closure is closed and bounded, so it's compact, and there are two points that realize the diameter. (Call these x and y.) If you take the closed disc for which this diameter is the diameter of the disc, I claim it would be such a smallest closed disc.
First if you had a smaller closed disc, then it's subset, F, would have a bigger diameter, a contradiction. Second, if there were points outside of this disc in F, you could use triangle inequality to argue that that point along with either x or y form a longer diameter in F.
answered Jan 3 at 21:17
GaffneyGaffney
1,403712
answered Jan 3 at 21:17
GaffneyGaffney
1,403712
answered Jan 3 at 21:17
GaffneyGaffney
1,403712
answered Jan 3 at 21:17
GaffneyGaffney
1,403712
1,403712
add a comment |
add a comment |
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$begingroup$
What's your definition of bounded? Is your problem that there will be A disc or that there will be a MINIMAL-radius disc?
$endgroup$
– Gaffney
Jan 3 at 21:02
$begingroup$
The minimal part. In other words why is there a circle which realizes the inf of radii of enclosing circles
$endgroup$
– Alex
Jan 3 at 21:05