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Prob. 4, Sec. 27, in Munkres' TOPOLOGY, 2nd ed: Any connected metric space having more than one point is uncountable
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Here is Prob. 4, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:
Show that a connected metric space having more than one point is uncountable.
Here is a solution. Although I do understand the proof at this URL [The gist of that proof is the fact that no finite or countably infinite subset of $[0, +infty)$ can be connected in the usual space $mathbbR$.], I'd like to attempt the following.
My Attempt:
Let $X$ be a set having more than one point. Suppose that $(X, d)$ is a metric space such that the set $X$ is either finite or countable.
Case 1.
If $X$ is finite, then we can suppose that $X = left x_1, ldots, x_n right$, where $n > 1$. Then, for each $j = 1, ldots, n$, let us put
$$ r_j colon= min left d left( x_i, x_j right) colon i = 1, ldots, n, i neq j right. tag1 $$
Then the open balls
$$ B_d left( x_j, r_j right) colon= left x in X colon d left( x, x_j right) < r_j right, $$
for $j = 1, ldots, n$, are open sets in $X$.
In fact, we also have
$$ B_d left( x_j, r_j right) = left x_j right, tag2 $$
because of our choice of $r_j$ in (1) above, for each $j = 1, ldots, n$.
So a separation (also called disconnection) of $X$ is given by
$$ X = C cup D, $$
where
$$ C colon= B_d left( x_1, r_1 right) qquad mbox and qquad D colon= bigcup_j=2^n B_d left( x_j, r_j right). $$
Thus $X$ is not connected.
Is my logic correct?
Case 2.
If $X$ is countably infinite, then suppose that $X$ has points $x_1, x_2, x_3, ldots$. That is, suppose
$$ X = left x_1, x_2, x_3, ldots right. $$
Then, as in Case 1, we can show that every finite subset of $X$ having more than one points is not connected.
Am I right?
Can we show from here that $X$ is not connected?
general-topology analysis proof-verification metric-spaces connectedness
edited 56 mins ago
Moo
4,7733920
asked 1 hour ago
Saaqib Mahmood
7,41442172
add a comment |Â
up vote
2
down vote
favorite
Here is Prob. 4, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:
Show that a connected metric space having more than one point is uncountable.
Here is a solution. Although I do understand the proof at this URL [The gist of that proof is the fact that no finite or countably infinite subset of $[0, +infty)$ can be connected in the usual space $mathbbR$.], I'd like to attempt the following.
My Attempt:
Let $X$ be a set having more than one point. Suppose that $(X, d)$ is a metric space such that the set $X$ is either finite or countable.
Case 1.
If $X$ is finite, then we can suppose that $X = left x_1, ldots, x_n right$, where $n > 1$. Then, for each $j = 1, ldots, n$, let us put
$$ r_j colon= min left d left( x_i, x_j right) colon i = 1, ldots, n, i neq j right. tag1 $$
Then the open balls
$$ B_d left( x_j, r_j right) colon= left x in X colon d left( x, x_j right) < r_j right, $$
for $j = 1, ldots, n$, are open sets in $X$.
In fact, we also have
$$ B_d left( x_j, r_j right) = left x_j right, tag2 $$
because of our choice of $r_j$ in (1) above, for each $j = 1, ldots, n$.
So a separation (also called disconnection) of $X$ is given by
$$ X = C cup D, $$
where
$$ C colon= B_d left( x_1, r_1 right) qquad mbox and qquad D colon= bigcup_j=2^n B_d left( x_j, r_j right). $$
Thus $X$ is not connected.
Is my logic correct?
Case 2.
If $X$ is countably infinite, then suppose that $X$ has points $x_1, x_2, x_3, ldots$. That is, suppose
$$ X = left x_1, x_2, x_3, ldots right. $$
Then, as in Case 1, we can show that every finite subset of $X$ having more than one points is not connected.
Am I right?
Can we show from here that $X$ is not connected?
general-topology analysis proof-verification metric-spaces connectedness
edited 56 mins ago
Moo
4,7733920
asked 1 hour ago
Saaqib Mahmood
7,41442172
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Here is Prob. 4, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:
Show that a connected metric space having more than one point is uncountable.
Here is a solution. Although I do understand the proof at this URL [The gist of that proof is the fact that no finite or countably infinite subset of $[0, +infty)$ can be connected in the usual space $mathbbR$.], I'd like to attempt the following.
My Attempt:
Let $X$ be a set having more than one point. Suppose that $(X, d)$ is a metric space such that the set $X$ is either finite or countable.
Case 1.
If $X$ is finite, then we can suppose that $X = left x_1, ldots, x_n right$, where $n > 1$. Then, for each $j = 1, ldots, n$, let us put
$$ r_j colon= min left d left( x_i, x_j right) colon i = 1, ldots, n, i neq j right. tag1 $$
Then the open balls
$$ B_d left( x_j, r_j right) colon= left x in X colon d left( x, x_j right) < r_j right, $$
for $j = 1, ldots, n$, are open sets in $X$.
In fact, we also have
$$ B_d left( x_j, r_j right) = left x_j right, tag2 $$
because of our choice of $r_j$ in (1) above, for each $j = 1, ldots, n$.
So a separation (also called disconnection) of $X$ is given by
$$ X = C cup D, $$
where
$$ C colon= B_d left( x_1, r_1 right) qquad mbox and qquad D colon= bigcup_j=2^n B_d left( x_j, r_j right). $$
Thus $X$ is not connected.
Is my logic correct?
Case 2.
If $X$ is countably infinite, then suppose that $X$ has points $x_1, x_2, x_3, ldots$. That is, suppose
$$ X = left x_1, x_2, x_3, ldots right. $$
Then, as in Case 1, we can show that every finite subset of $X$ having more than one points is not connected.
Am I right?
Can we show from here that $X$ is not connected?
general-topology analysis proof-verification metric-spaces connectedness
edited 56 mins ago
Moo
4,7733920
asked 1 hour ago
Saaqib Mahmood
7,41442172
Here is Prob. 4, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:
Show that a connected metric space having more than one point is uncountable.
Here is a solution. Although I do understand the proof at this URL [The gist of that proof is the fact that no finite or countably infinite subset of $[0, +infty)$ can be connected in the usual space $mathbbR$.], I'd like to attempt the following.
My Attempt:
Let $X$ be a set having more than one point. Suppose that $(X, d)$ is a metric space such that the set $X$ is either finite or countable.
Case 1.
If $X$ is finite, then we can suppose that $X = left x_1, ldots, x_n right$, where $n > 1$. Then, for each $j = 1, ldots, n$, let us put
$$ r_j colon= min left d left( x_i, x_j right) colon i = 1, ldots, n, i neq j right. tag1 $$
Then the open balls
$$ B_d left( x_j, r_j right) colon= left x in X colon d left( x, x_j right) < r_j right, $$
for $j = 1, ldots, n$, are open sets in $X$.
In fact, we also have
$$ B_d left( x_j, r_j right) = left x_j right, tag2 $$
because of our choice of $r_j$ in (1) above, for each $j = 1, ldots, n$.
So a separation (also called disconnection) of $X$ is given by
$$ X = C cup D, $$
where
$$ C colon= B_d left( x_1, r_1 right) qquad mbox and qquad D colon= bigcup_j=2^n B_d left( x_j, r_j right). $$
Thus $X$ is not connected.
Is my logic correct?
Case 2.
If $X$ is countably infinite, then suppose that $X$ has points $x_1, x_2, x_3, ldots$. That is, suppose
$$ X = left x_1, x_2, x_3, ldots right. $$
Then, as in Case 1, we can show that every finite subset of $X$ having more than one points is not connected.
Am I right?
Can we show from here that $X$ is not connected?
general-topology analysis proof-verification metric-spaces connectedness
general-topology analysis proof-verification metric-spaces connectedness
edited 56 mins ago
Moo
4,7733920
asked 1 hour ago
Saaqib Mahmood
7,41442172
edited 56 mins ago
Moo
4,7733920
asked 1 hour ago
Saaqib Mahmood
7,41442172
edited 56 mins ago
Moo
4,7733920
edited 56 mins ago
Moo
4,7733920
edited 56 mins ago
Moo
4,7733920
4,7733920
asked 1 hour ago
Saaqib Mahmood
7,41442172
asked 1 hour ago
Saaqib Mahmood
7,41442172
asked 1 hour ago
Saaqib Mahmood
7,41442172
7,41442172
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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up vote
3
down vote
Here is an argument that does not require Urysohn's Lemma: fix $x in X$ and define $f(y)=d(x,y)$. Then $f:X to [0,infty)$ is continuous, $f(x)=0$ and $f(y) >0$ for some $y$. Since the range is an interval it is uncountable and so is $X$.
answered 1 hour ago
Kavi Rama Murthy
28.9k31439
add a comment |Â
up vote
3
down vote
Assume distinct $a,b$. By Urysohn's thereom there's a
continuous $f:X to [0,1]$, with $f(a) = 0, f(b) = 1$.
Since $X$ is connected, so is $f(X)$. Thus $f(X) = [0,1]$.
Whence $X$ cannot be countable.
edited 57 mins ago
Moo
4,7733920
answered 1 hour ago
William Elliot
5,6252517
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Here is an argument that does not require Urysohn's Lemma: fix $x in X$ and define $f(y)=d(x,y)$. Then $f:X to [0,infty)$ is continuous, $f(x)=0$ and $f(y) >0$ for some $y$. Since the range is an interval it is uncountable and so is $X$.
answered 1 hour ago
Kavi Rama Murthy
28.9k31439
add a comment |Â
up vote
3
down vote
Here is an argument that does not require Urysohn's Lemma: fix $x in X$ and define $f(y)=d(x,y)$. Then $f:X to [0,infty)$ is continuous, $f(x)=0$ and $f(y) >0$ for some $y$. Since the range is an interval it is uncountable and so is $X$.
answered 1 hour ago
Kavi Rama Murthy
28.9k31439
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Here is an argument that does not require Urysohn's Lemma: fix $x in X$ and define $f(y)=d(x,y)$. Then $f:X to [0,infty)$ is continuous, $f(x)=0$ and $f(y) >0$ for some $y$. Since the range is an interval it is uncountable and so is $X$.
answered 1 hour ago
Kavi Rama Murthy
28.9k31439
Here is an argument that does not require Urysohn's Lemma: fix $x in X$ and define $f(y)=d(x,y)$. Then $f:X to [0,infty)$ is continuous, $f(x)=0$ and $f(y) >0$ for some $y$. Since the range is an interval it is uncountable and so is $X$.
answered 1 hour ago
Kavi Rama Murthy
28.9k31439
answered 1 hour ago
Kavi Rama Murthy
28.9k31439
answered 1 hour ago
Kavi Rama Murthy
28.9k31439
answered 1 hour ago
Kavi Rama Murthy
28.9k31439
28.9k31439
add a comment |Â
add a comment |Â
up vote
3
down vote
Assume distinct $a,b$. By Urysohn's thereom there's a
continuous $f:X to [0,1]$, with $f(a) = 0, f(b) = 1$.
Since $X$ is connected, so is $f(X)$. Thus $f(X) = [0,1]$.
Whence $X$ cannot be countable.
edited 57 mins ago
Moo
4,7733920
answered 1 hour ago
William Elliot
5,6252517
add a comment |Â
up vote
3
down vote
Assume distinct $a,b$. By Urysohn's thereom there's a
continuous $f:X to [0,1]$, with $f(a) = 0, f(b) = 1$.
Since $X$ is connected, so is $f(X)$. Thus $f(X) = [0,1]$.
Whence $X$ cannot be countable.
edited 57 mins ago
Moo
4,7733920
answered 1 hour ago
William Elliot
5,6252517
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Assume distinct $a,b$. By Urysohn's thereom there's a
continuous $f:X to [0,1]$, with $f(a) = 0, f(b) = 1$.
Since $X$ is connected, so is $f(X)$. Thus $f(X) = [0,1]$.
Whence $X$ cannot be countable.
edited 57 mins ago
Moo
4,7733920
answered 1 hour ago
William Elliot
5,6252517
Assume distinct $a,b$. By Urysohn's thereom there's a
continuous $f:X to [0,1]$, with $f(a) = 0, f(b) = 1$.
Since $X$ is connected, so is $f(X)$. Thus $f(X) = [0,1]$.
Whence $X$ cannot be countable.
edited 57 mins ago
Moo
4,7733920
answered 1 hour ago
William Elliot
5,6252517
edited 57 mins ago
Moo
4,7733920
edited 57 mins ago
Moo
4,7733920
edited 57 mins ago
Moo
4,7733920
4,7733920
answered 1 hour ago
William Elliot
5,6252517
answered 1 hour ago
William Elliot
5,6252517
answered 1 hour ago
William Elliot
5,6252517
5,6252517
add a comment |Â
add a comment |Â
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