Mixing
Prob. 2(b), Sec. 25, in Munkres' TOPOLOGY, 2nd ed: The iff-condition for two points to be in the same component of $mathbbR^omega$
Clash Royale CLAN TAG#URR8PPP
up vote
12
down vote
favorite
Here is Prob. 2 (b), Sec. 25, in the book Topology by James R. Munkres, 2nd edition:
Consider $mathbbR^omega$ in the uniform topology. Show that $mathbfx$ and $mathbfy$ lie in the same component of $mathbbR^omega$ if and only if the sequence $$ mathbfx - mathbfy = left( x_1 - y_1, x_2 - y_2, ldots right) $$
is bounded. [Hint: It suffices to consider the case where $mathbfy = mathbf0$.]
My Attempt:
First of all, we note that $mathbbR^omega$ denotes the set of all the (infinite) sequences of real numbers; that is, $mathbbR^omega$ denotes the countably infinite Cartesian product
$$ mathbbR times mathbbR times cdots. $$
And, the uniform topology on $mathbbR^omega$ is the one induced by the so-called uniform metric on $mathbbR^omega$, which is given by
$$ barrho( mathbfx, mathbfy ) colon= sup left min left leftlvert x_n - y_n rightrvert, 1 right colon n in mathbbN right $$
for all $mathbfx colon= left( x_n right)_n in mathbbN$ and $mathbfy colon= left( y_n right)_n in mathbbN$ in $mathbbR^omega$.
Now let $mathbfa colon= left( a_n right)_n in mathbbN$ and $mathbfb colon= left( b_n right)_n in mathbbN$ be any two points in $mathbbR^omega$ such that the sequence $$ mathbfa - mathbfb = left( a_n - b_n right)_n in mathbbN$$
is bounded, that is, such that there exists a positive real number $r$ such that
$$ leftlvert a_n - b_n rightrvert leq r mbox for all n in mathbbN. tag1 $$
Now let $f colon [ 0, 1 ] to mathbbR^omega$ be the map defined by
$$ f(t) colon= mathbfa + t(mathbfb- mathbfa ) = big( a_n + t left( b_n - a_n right) big)_n in mathbbN mbox for all t in [0, 1]. $$
Then for any points $s, t in [0, 1]$, we note that
$$
beginalign
barrho( f(s), f(t) ) &= sup left min left biglvert big( a_n + s left( b_n - a_n right) big) - big( a_n + t left( b_n - a_n right) big) bigrvert, , 1 right colon n in mathbbN right \
&= sup left min left lvert s-t rvert cdot leftlvert b_n - a_n rightrvert, , 1 right colon n in mathbbN right \
&leq sup left lvert s-t rvert cdot leftlvert b_n - a_n rightrvert colon n in mathbbN right \
&= lvert s-t rvert cdot sup left leftlvert b_n - a_n rightrvert colon n in mathbbN right \
&leq lvert s-t rvert r qquad mbox [ using (1) above ].
endalign
$$
So, given any real number $varepsilon > 0$, if we choose our real number $delta$ such that $$ 0 < delta < fracvarepsilonr, $$
then we find that
$$ barrho( f(s), f(t) ) < varepsilon $$
holds for any pair of points $s, t in [0, 1]$ for which $$ lvert s-t rvert < delta. $$
Thus the map $f$ is continuous (in fact uniformly continuous). Moreover, $f(0) = mathbfa$ and $f(1) = mathbfb$. Therefore $f$ is a path in the uniform space $mathbbR^omega$ joining the points $mathbfa$ and $mathbfb$. So $mathbfa$ and $mathbfb$ lie in the same path component of $mathbbR^omega$. And since each path component of any topological space is contained in a component of that space, we can conclude that $mathbfa$ and $mathbfb$ lie in the same component of $mathbbR^omega$.
Am I right? Can we give an independent proof of the above without having recourse to path components?
Now suppose that the sequence $mathbfa - mathbfb$ is unbounded. Then, for any natural number $k$, there exists a natural number $n_k$ such that
$$ leftlvert a_n_k - b_n_k rightrvert > k. $$
Now let $U$ be the set of all the points $mathbfx colon= left( x_n right)_n in mathbbN$ in $mathbbR^omega$ such that the sequence $mathbfa - mathbfx$ is bounded. Then $mathbfa in U$.
Moreover, If $mathbfu colon= left( u_n right)_n in mathbbN in U$, then the sequence $left( a_n - u_n right)_n in mathbbN$ is bounded; that is, there exists a positive real number $r_mathbfu$ such that
$$ leftlvert a_n - u_n rightrvert < r_mathbfu mbox for all n in mathbbN. $$
So if $varepsilon in (0, 1)$ and if $mathbfx$ is any point of $mathbbR^omega$ such that $$ barrho( mathbfx, mathbfu ) < varepsilon, $$
then, for each $n in mathbbN$, we have
$$
minbig leftlvert x_n - u_n rightrvert, 1 big leq barrho( mathbfx, mathbfu ) < varepsilon < 1,
$$
and so
$$
minbig leftlvert x_n - u_n rightrvert, 1 big = leftlvert x_n - u_n rightrvert,
$$
and thus
$$ leftlvert x_n - u_n rightrvert < varepsilon. $$
Then, for any $n in mathbbN$, we obtain
$$
beginalign
leftlvert x_n - a_n rightrvert &leq leftlvert x_n - u_n rightrvert + leftlvert u_n - a_n rightrvert \
&< varepsilon + r_mathbfu.
endalign
$$
thus showing that $mathbfx in U$ also. Therefore $U$ is open in the uniform space $mathbbR^omega$.
Now the point $mathbfb$ is in the set $mathbbR^omega setminus U$.
Let $mathbfv$ be any point of $mathbbR^omega setminus U$. Then the sequence $left( v_n - a_n right)_n in mathbbN$ is unbounded; so for any natural number $k$ there exists a natural number $m_k$ such that
$$ leftlvert v_m_k - a_m_k rightrvert > k. $$
Now if $varepsilon in (0, 1)$, and if $mathbfy$ is any point of $mathbbR^omega$ such that
$$ barrho( mathbfy, mathbfv ) < varepsilon, $$
then as before we must have
$$ leftlvert y_n - v_n rightrvert < varepsilon $$
for every natural number $n$. Now if $lambda$ is an arbitrary real number and if $k$ is a natural number such that $k > lambda + varepsilon$, then we find that
$$
beginalign
leftlvert y_m_k - a_m_k rightrvert &geq leftlvert v_m_k - a_m_k rightrvert - leftlvert y_m_k - v_m_k rightrvert \
&> k - varepsilon \
&> lambda.
endalign
$$
Thus corresponding to any real number $lambda$, no matter how large, we can find a natural number $m_k$ such that
$$ leftlvert y_m_k - a_m_k rightrvert > lambda. $$
Thus the sequence $left( y_n - a_n right)_n in mathbbN$ is unbounded, which shows that $mathbfy$ is in $mathbbR^omega setminus U$. Therefore the set $mathbbR^omega setminus U$ is also open in the uniform metric space $mathbbR^omega$.
Thus the sets $U$ and $mathbbR^omega setminus U$ constitute a separation of the uniform metric space $mathbbR^omega$. Moreover, these two sets are both open and closed in $mathbbR^omega$ with the uniform topology.
Finally, if $mathbfx$ and $mathbfy$ are any two points of $U$. then the sequences $mathbfx - mathbfa$ and $mathbfy - mathbfa$ are both bounded; so the sequence $mathbfx - mathbfy = ( mathbfx - mathbfa ) - ( mathbfy - mathbfa )$ is bounded as well. Thus $mathbfx$ and $mathbfy$ are in the same (path) component of the uniform metric space $mathbbR^omega$, as has been shown above. Thus the set $U$ is (path) connected.
Similarly, the set $V$ of all the points $mathbfv$ in $mathbbR^omega$ such that $mathbfv - mathbfb$ is bounded is (path) connected. Moreover, $mathbfb$ is in $V$ but $mathbfa$ is not in $V$. Also $V$ is both open and closed in the uniform metric space $mathbbR^omega$, just as $U$ has been shown to be both open and closed.
Moreover, the sets $U$ and $V$ are disjoint.
I'm getting lost now!
What next? How to proceed from here? Or, is there anywhere I've gone wrong?
P.S.:
I think I've just managed to hit upon the right trick!
Now as the sets $U$ and $mathbbR^omega setminus U$ form a separation of the uniform metric space $mathbbR^omega$, so any connected subspace lies in either $U$ or $mathbbR^omega setminus U$, but not both. [Refer to Lemma 23.2 in Munkres.]
As $mathbfa in U$, so the component containing $mathbfa$, which is of course a connected subspace of $mathbbR^omega$, must also lie in $U$. And, as $mathbfb$ is in $mathbbR^omega setminus U$, so the component containing $mathbfb$ lies in $mathbbR^omega setminus U$. Therefore the points $mathbfa$ and $mathbfb$ lie in different components of $mathbbR^omega$ in the uniform topology if the sequence $mathbfa - mathbfb$ is unbounded.
Is there any flaw in my reasoning?
general-topology proof-verification connectedness alternative-proof path-connected
asked Jul 27 at 19:50
Saaqib Mahmood
7,11042169
add a comment |Â
up vote
12
down vote
favorite
Here is Prob. 2 (b), Sec. 25, in the book Topology by James R. Munkres, 2nd edition:
Consider $mathbbR^omega$ in the uniform topology. Show that $mathbfx$ and $mathbfy$ lie in the same component of $mathbbR^omega$ if and only if the sequence $$ mathbfx - mathbfy = left( x_1 - y_1, x_2 - y_2, ldots right) $$
is bounded. [Hint: It suffices to consider the case where $mathbfy = mathbf0$.]
My Attempt:
First of all, we note that $mathbbR^omega$ denotes the set of all the (infinite) sequences of real numbers; that is, $mathbbR^omega$ denotes the countably infinite Cartesian product
$$ mathbbR times mathbbR times cdots. $$
And, the uniform topology on $mathbbR^omega$ is the one induced by the so-called uniform metric on $mathbbR^omega$, which is given by
$$ barrho( mathbfx, mathbfy ) colon= sup left min left leftlvert x_n - y_n rightrvert, 1 right colon n in mathbbN right $$
for all $mathbfx colon= left( x_n right)_n in mathbbN$ and $mathbfy colon= left( y_n right)_n in mathbbN$ in $mathbbR^omega$.
Now let $mathbfa colon= left( a_n right)_n in mathbbN$ and $mathbfb colon= left( b_n right)_n in mathbbN$ be any two points in $mathbbR^omega$ such that the sequence $$ mathbfa - mathbfb = left( a_n - b_n right)_n in mathbbN$$
is bounded, that is, such that there exists a positive real number $r$ such that
$$ leftlvert a_n - b_n rightrvert leq r mbox for all n in mathbbN. tag1 $$
Now let $f colon [ 0, 1 ] to mathbbR^omega$ be the map defined by
$$ f(t) colon= mathbfa + t(mathbfb- mathbfa ) = big( a_n + t left( b_n - a_n right) big)_n in mathbbN mbox for all t in [0, 1]. $$
Then for any points $s, t in [0, 1]$, we note that
$$
beginalign
barrho( f(s), f(t) ) &= sup left min left biglvert big( a_n + s left( b_n - a_n right) big) - big( a_n + t left( b_n - a_n right) big) bigrvert, , 1 right colon n in mathbbN right \
&= sup left min left lvert s-t rvert cdot leftlvert b_n - a_n rightrvert, , 1 right colon n in mathbbN right \
&leq sup left lvert s-t rvert cdot leftlvert b_n - a_n rightrvert colon n in mathbbN right \
&= lvert s-t rvert cdot sup left leftlvert b_n - a_n rightrvert colon n in mathbbN right \
&leq lvert s-t rvert r qquad mbox [ using (1) above ].
endalign
$$
So, given any real number $varepsilon > 0$, if we choose our real number $delta$ such that $$ 0 < delta < fracvarepsilonr, $$
then we find that
$$ barrho( f(s), f(t) ) < varepsilon $$
holds for any pair of points $s, t in [0, 1]$ for which $$ lvert s-t rvert < delta. $$
Thus the map $f$ is continuous (in fact uniformly continuous). Moreover, $f(0) = mathbfa$ and $f(1) = mathbfb$. Therefore $f$ is a path in the uniform space $mathbbR^omega$ joining the points $mathbfa$ and $mathbfb$. So $mathbfa$ and $mathbfb$ lie in the same path component of $mathbbR^omega$. And since each path component of any topological space is contained in a component of that space, we can conclude that $mathbfa$ and $mathbfb$ lie in the same component of $mathbbR^omega$.
Am I right? Can we give an independent proof of the above without having recourse to path components?
Now suppose that the sequence $mathbfa - mathbfb$ is unbounded. Then, for any natural number $k$, there exists a natural number $n_k$ such that
$$ leftlvert a_n_k - b_n_k rightrvert > k. $$
Now let $U$ be the set of all the points $mathbfx colon= left( x_n right)_n in mathbbN$ in $mathbbR^omega$ such that the sequence $mathbfa - mathbfx$ is bounded. Then $mathbfa in U$.
Moreover, If $mathbfu colon= left( u_n right)_n in mathbbN in U$, then the sequence $left( a_n - u_n right)_n in mathbbN$ is bounded; that is, there exists a positive real number $r_mathbfu$ such that
$$ leftlvert a_n - u_n rightrvert < r_mathbfu mbox for all n in mathbbN. $$
So if $varepsilon in (0, 1)$ and if $mathbfx$ is any point of $mathbbR^omega$ such that $$ barrho( mathbfx, mathbfu ) < varepsilon, $$
then, for each $n in mathbbN$, we have
$$
minbig leftlvert x_n - u_n rightrvert, 1 big leq barrho( mathbfx, mathbfu ) < varepsilon < 1,
$$
and so
$$
minbig leftlvert x_n - u_n rightrvert, 1 big = leftlvert x_n - u_n rightrvert,
$$
and thus
$$ leftlvert x_n - u_n rightrvert < varepsilon. $$
Then, for any $n in mathbbN$, we obtain
$$
beginalign
leftlvert x_n - a_n rightrvert &leq leftlvert x_n - u_n rightrvert + leftlvert u_n - a_n rightrvert \
&< varepsilon + r_mathbfu.
endalign
$$
thus showing that $mathbfx in U$ also. Therefore $U$ is open in the uniform space $mathbbR^omega$.
Now the point $mathbfb$ is in the set $mathbbR^omega setminus U$.
Let $mathbfv$ be any point of $mathbbR^omega setminus U$. Then the sequence $left( v_n - a_n right)_n in mathbbN$ is unbounded; so for any natural number $k$ there exists a natural number $m_k$ such that
$$ leftlvert v_m_k - a_m_k rightrvert > k. $$
Now if $varepsilon in (0, 1)$, and if $mathbfy$ is any point of $mathbbR^omega$ such that
$$ barrho( mathbfy, mathbfv ) < varepsilon, $$
then as before we must have
$$ leftlvert y_n - v_n rightrvert < varepsilon $$
for every natural number $n$. Now if $lambda$ is an arbitrary real number and if $k$ is a natural number such that $k > lambda + varepsilon$, then we find that
$$
beginalign
leftlvert y_m_k - a_m_k rightrvert &geq leftlvert v_m_k - a_m_k rightrvert - leftlvert y_m_k - v_m_k rightrvert \
&> k - varepsilon \
&> lambda.
endalign
$$
Thus corresponding to any real number $lambda$, no matter how large, we can find a natural number $m_k$ such that
$$ leftlvert y_m_k - a_m_k rightrvert > lambda. $$
Thus the sequence $left( y_n - a_n right)_n in mathbbN$ is unbounded, which shows that $mathbfy$ is in $mathbbR^omega setminus U$. Therefore the set $mathbbR^omega setminus U$ is also open in the uniform metric space $mathbbR^omega$.
Thus the sets $U$ and $mathbbR^omega setminus U$ constitute a separation of the uniform metric space $mathbbR^omega$. Moreover, these two sets are both open and closed in $mathbbR^omega$ with the uniform topology.
Finally, if $mathbfx$ and $mathbfy$ are any two points of $U$. then the sequences $mathbfx - mathbfa$ and $mathbfy - mathbfa$ are both bounded; so the sequence $mathbfx - mathbfy = ( mathbfx - mathbfa ) - ( mathbfy - mathbfa )$ is bounded as well. Thus $mathbfx$ and $mathbfy$ are in the same (path) component of the uniform metric space $mathbbR^omega$, as has been shown above. Thus the set $U$ is (path) connected.
Similarly, the set $V$ of all the points $mathbfv$ in $mathbbR^omega$ such that $mathbfv - mathbfb$ is bounded is (path) connected. Moreover, $mathbfb$ is in $V$ but $mathbfa$ is not in $V$. Also $V$ is both open and closed in the uniform metric space $mathbbR^omega$, just as $U$ has been shown to be both open and closed.
Moreover, the sets $U$ and $V$ are disjoint.
I'm getting lost now!
What next? How to proceed from here? Or, is there anywhere I've gone wrong?
P.S.:
I think I've just managed to hit upon the right trick!
Now as the sets $U$ and $mathbbR^omega setminus U$ form a separation of the uniform metric space $mathbbR^omega$, so any connected subspace lies in either $U$ or $mathbbR^omega setminus U$, but not both. [Refer to Lemma 23.2 in Munkres.]
As $mathbfa in U$, so the component containing $mathbfa$, which is of course a connected subspace of $mathbbR^omega$, must also lie in $U$. And, as $mathbfb$ is in $mathbbR^omega setminus U$, so the component containing $mathbfb$ lies in $mathbbR^omega setminus U$. Therefore the points $mathbfa$ and $mathbfb$ lie in different components of $mathbbR^omega$ in the uniform topology if the sequence $mathbfa - mathbfb$ is unbounded.
Is there any flaw in my reasoning?
general-topology proof-verification connectedness alternative-proof path-connected
asked Jul 27 at 19:50
Saaqib Mahmood
7,11042169
1
Does your notational distinction between $mathbf R$ and $mathbb R$ carry any significance? If so, it should be explained; if not, it should be removed.
– joriki
Jul 28 at 3:00
@joriki I write $mathbbR$ to denote the set of real numbers. I'll just edit my post to omit $mathbfR$ whereever I've used it. Thanks for the comment.
– Saaqib Mahmood
Jul 28 at 11:38
See also math.stackexchange.com/q/2730869.
– Paul Frost
Jul 28 at 13:51
add a comment |Â
up vote
12
down vote
favorite
up vote
12
down vote
favorite
Here is Prob. 2 (b), Sec. 25, in the book Topology by James R. Munkres, 2nd edition:
Consider $mathbbR^omega$ in the uniform topology. Show that $mathbfx$ and $mathbfy$ lie in the same component of $mathbbR^omega$ if and only if the sequence $$ mathbfx - mathbfy = left( x_1 - y_1, x_2 - y_2, ldots right) $$
is bounded. [Hint: It suffices to consider the case where $mathbfy = mathbf0$.]
My Attempt:
First of all, we note that $mathbbR^omega$ denotes the set of all the (infinite) sequences of real numbers; that is, $mathbbR^omega$ denotes the countably infinite Cartesian product
$$ mathbbR times mathbbR times cdots. $$
And, the uniform topology on $mathbbR^omega$ is the one induced by the so-called uniform metric on $mathbbR^omega$, which is given by
$$ barrho( mathbfx, mathbfy ) colon= sup left min left leftlvert x_n - y_n rightrvert, 1 right colon n in mathbbN right $$
for all $mathbfx colon= left( x_n right)_n in mathbbN$ and $mathbfy colon= left( y_n right)_n in mathbbN$ in $mathbbR^omega$.
Now let $mathbfa colon= left( a_n right)_n in mathbbN$ and $mathbfb colon= left( b_n right)_n in mathbbN$ be any two points in $mathbbR^omega$ such that the sequence $$ mathbfa - mathbfb = left( a_n - b_n right)_n in mathbbN$$
is bounded, that is, such that there exists a positive real number $r$ such that
$$ leftlvert a_n - b_n rightrvert leq r mbox for all n in mathbbN. tag1 $$
Now let $f colon [ 0, 1 ] to mathbbR^omega$ be the map defined by
$$ f(t) colon= mathbfa + t(mathbfb- mathbfa ) = big( a_n + t left( b_n - a_n right) big)_n in mathbbN mbox for all t in [0, 1]. $$
Then for any points $s, t in [0, 1]$, we note that
$$
beginalign
barrho( f(s), f(t) ) &= sup left min left biglvert big( a_n + s left( b_n - a_n right) big) - big( a_n + t left( b_n - a_n right) big) bigrvert, , 1 right colon n in mathbbN right \
&= sup left min left lvert s-t rvert cdot leftlvert b_n - a_n rightrvert, , 1 right colon n in mathbbN right \
&leq sup left lvert s-t rvert cdot leftlvert b_n - a_n rightrvert colon n in mathbbN right \
&= lvert s-t rvert cdot sup left leftlvert b_n - a_n rightrvert colon n in mathbbN right \
&leq lvert s-t rvert r qquad mbox [ using (1) above ].
endalign
$$
So, given any real number $varepsilon > 0$, if we choose our real number $delta$ such that $$ 0 < delta < fracvarepsilonr, $$
then we find that
$$ barrho( f(s), f(t) ) < varepsilon $$
holds for any pair of points $s, t in [0, 1]$ for which $$ lvert s-t rvert < delta. $$
Thus the map $f$ is continuous (in fact uniformly continuous). Moreover, $f(0) = mathbfa$ and $f(1) = mathbfb$. Therefore $f$ is a path in the uniform space $mathbbR^omega$ joining the points $mathbfa$ and $mathbfb$. So $mathbfa$ and $mathbfb$ lie in the same path component of $mathbbR^omega$. And since each path component of any topological space is contained in a component of that space, we can conclude that $mathbfa$ and $mathbfb$ lie in the same component of $mathbbR^omega$.
Am I right? Can we give an independent proof of the above without having recourse to path components?
Now suppose that the sequence $mathbfa - mathbfb$ is unbounded. Then, for any natural number $k$, there exists a natural number $n_k$ such that
$$ leftlvert a_n_k - b_n_k rightrvert > k. $$
Now let $U$ be the set of all the points $mathbfx colon= left( x_n right)_n in mathbbN$ in $mathbbR^omega$ such that the sequence $mathbfa - mathbfx$ is bounded. Then $mathbfa in U$.
Moreover, If $mathbfu colon= left( u_n right)_n in mathbbN in U$, then the sequence $left( a_n - u_n right)_n in mathbbN$ is bounded; that is, there exists a positive real number $r_mathbfu$ such that
$$ leftlvert a_n - u_n rightrvert < r_mathbfu mbox for all n in mathbbN. $$
So if $varepsilon in (0, 1)$ and if $mathbfx$ is any point of $mathbbR^omega$ such that $$ barrho( mathbfx, mathbfu ) < varepsilon, $$
then, for each $n in mathbbN$, we have
$$
minbig leftlvert x_n - u_n rightrvert, 1 big leq barrho( mathbfx, mathbfu ) < varepsilon < 1,
$$
and so
$$
minbig leftlvert x_n - u_n rightrvert, 1 big = leftlvert x_n - u_n rightrvert,
$$
and thus
$$ leftlvert x_n - u_n rightrvert < varepsilon. $$
Then, for any $n in mathbbN$, we obtain
$$
beginalign
leftlvert x_n - a_n rightrvert &leq leftlvert x_n - u_n rightrvert + leftlvert u_n - a_n rightrvert \
&< varepsilon + r_mathbfu.
endalign
$$
thus showing that $mathbfx in U$ also. Therefore $U$ is open in the uniform space $mathbbR^omega$.
Now the point $mathbfb$ is in the set $mathbbR^omega setminus U$.
Let $mathbfv$ be any point of $mathbbR^omega setminus U$. Then the sequence $left( v_n - a_n right)_n in mathbbN$ is unbounded; so for any natural number $k$ there exists a natural number $m_k$ such that
$$ leftlvert v_m_k - a_m_k rightrvert > k. $$
Now if $varepsilon in (0, 1)$, and if $mathbfy$ is any point of $mathbbR^omega$ such that
$$ barrho( mathbfy, mathbfv ) < varepsilon, $$
then as before we must have
$$ leftlvert y_n - v_n rightrvert < varepsilon $$
for every natural number $n$. Now if $lambda$ is an arbitrary real number and if $k$ is a natural number such that $k > lambda + varepsilon$, then we find that
$$
beginalign
leftlvert y_m_k - a_m_k rightrvert &geq leftlvert v_m_k - a_m_k rightrvert - leftlvert y_m_k - v_m_k rightrvert \
&> k - varepsilon \
&> lambda.
endalign
$$
Thus corresponding to any real number $lambda$, no matter how large, we can find a natural number $m_k$ such that
$$ leftlvert y_m_k - a_m_k rightrvert > lambda. $$
Thus the sequence $left( y_n - a_n right)_n in mathbbN$ is unbounded, which shows that $mathbfy$ is in $mathbbR^omega setminus U$. Therefore the set $mathbbR^omega setminus U$ is also open in the uniform metric space $mathbbR^omega$.
Thus the sets $U$ and $mathbbR^omega setminus U$ constitute a separation of the uniform metric space $mathbbR^omega$. Moreover, these two sets are both open and closed in $mathbbR^omega$ with the uniform topology.
Finally, if $mathbfx$ and $mathbfy$ are any two points of $U$. then the sequences $mathbfx - mathbfa$ and $mathbfy - mathbfa$ are both bounded; so the sequence $mathbfx - mathbfy = ( mathbfx - mathbfa ) - ( mathbfy - mathbfa )$ is bounded as well. Thus $mathbfx$ and $mathbfy$ are in the same (path) component of the uniform metric space $mathbbR^omega$, as has been shown above. Thus the set $U$ is (path) connected.
Similarly, the set $V$ of all the points $mathbfv$ in $mathbbR^omega$ such that $mathbfv - mathbfb$ is bounded is (path) connected. Moreover, $mathbfb$ is in $V$ but $mathbfa$ is not in $V$. Also $V$ is both open and closed in the uniform metric space $mathbbR^omega$, just as $U$ has been shown to be both open and closed.
Moreover, the sets $U$ and $V$ are disjoint.
I'm getting lost now!
What next? How to proceed from here? Or, is there anywhere I've gone wrong?
P.S.:
I think I've just managed to hit upon the right trick!
Now as the sets $U$ and $mathbbR^omega setminus U$ form a separation of the uniform metric space $mathbbR^omega$, so any connected subspace lies in either $U$ or $mathbbR^omega setminus U$, but not both. [Refer to Lemma 23.2 in Munkres.]
As $mathbfa in U$, so the component containing $mathbfa$, which is of course a connected subspace of $mathbbR^omega$, must also lie in $U$. And, as $mathbfb$ is in $mathbbR^omega setminus U$, so the component containing $mathbfb$ lies in $mathbbR^omega setminus U$. Therefore the points $mathbfa$ and $mathbfb$ lie in different components of $mathbbR^omega$ in the uniform topology if the sequence $mathbfa - mathbfb$ is unbounded.
Is there any flaw in my reasoning?
general-topology proof-verification connectedness alternative-proof path-connected
asked Jul 27 at 19:50
Saaqib Mahmood
7,11042169
Here is Prob. 2 (b), Sec. 25, in the book Topology by James R. Munkres, 2nd edition:
Consider $mathbbR^omega$ in the uniform topology. Show that $mathbfx$ and $mathbfy$ lie in the same component of $mathbbR^omega$ if and only if the sequence $$ mathbfx - mathbfy = left( x_1 - y_1, x_2 - y_2, ldots right) $$
is bounded. [Hint: It suffices to consider the case where $mathbfy = mathbf0$.]
My Attempt:
First of all, we note that $mathbbR^omega$ denotes the set of all the (infinite) sequences of real numbers; that is, $mathbbR^omega$ denotes the countably infinite Cartesian product
$$ mathbbR times mathbbR times cdots. $$
And, the uniform topology on $mathbbR^omega$ is the one induced by the so-called uniform metric on $mathbbR^omega$, which is given by
$$ barrho( mathbfx, mathbfy ) colon= sup left min left leftlvert x_n - y_n rightrvert, 1 right colon n in mathbbN right $$
for all $mathbfx colon= left( x_n right)_n in mathbbN$ and $mathbfy colon= left( y_n right)_n in mathbbN$ in $mathbbR^omega$.
Now let $mathbfa colon= left( a_n right)_n in mathbbN$ and $mathbfb colon= left( b_n right)_n in mathbbN$ be any two points in $mathbbR^omega$ such that the sequence $$ mathbfa - mathbfb = left( a_n - b_n right)_n in mathbbN$$
is bounded, that is, such that there exists a positive real number $r$ such that
$$ leftlvert a_n - b_n rightrvert leq r mbox for all n in mathbbN. tag1 $$
Now let $f colon [ 0, 1 ] to mathbbR^omega$ be the map defined by
$$ f(t) colon= mathbfa + t(mathbfb- mathbfa ) = big( a_n + t left( b_n - a_n right) big)_n in mathbbN mbox for all t in [0, 1]. $$
Then for any points $s, t in [0, 1]$, we note that
$$
beginalign
barrho( f(s), f(t) ) &= sup left min left biglvert big( a_n + s left( b_n - a_n right) big) - big( a_n + t left( b_n - a_n right) big) bigrvert, , 1 right colon n in mathbbN right \
&= sup left min left lvert s-t rvert cdot leftlvert b_n - a_n rightrvert, , 1 right colon n in mathbbN right \
&leq sup left lvert s-t rvert cdot leftlvert b_n - a_n rightrvert colon n in mathbbN right \
&= lvert s-t rvert cdot sup left leftlvert b_n - a_n rightrvert colon n in mathbbN right \
&leq lvert s-t rvert r qquad mbox [ using (1) above ].
endalign
$$
So, given any real number $varepsilon > 0$, if we choose our real number $delta$ such that $$ 0 < delta < fracvarepsilonr, $$
then we find that
$$ barrho( f(s), f(t) ) < varepsilon $$
holds for any pair of points $s, t in [0, 1]$ for which $$ lvert s-t rvert < delta. $$
Thus the map $f$ is continuous (in fact uniformly continuous). Moreover, $f(0) = mathbfa$ and $f(1) = mathbfb$. Therefore $f$ is a path in the uniform space $mathbbR^omega$ joining the points $mathbfa$ and $mathbfb$. So $mathbfa$ and $mathbfb$ lie in the same path component of $mathbbR^omega$. And since each path component of any topological space is contained in a component of that space, we can conclude that $mathbfa$ and $mathbfb$ lie in the same component of $mathbbR^omega$.
Am I right? Can we give an independent proof of the above without having recourse to path components?
Now suppose that the sequence $mathbfa - mathbfb$ is unbounded. Then, for any natural number $k$, there exists a natural number $n_k$ such that
$$ leftlvert a_n_k - b_n_k rightrvert > k. $$
Now let $U$ be the set of all the points $mathbfx colon= left( x_n right)_n in mathbbN$ in $mathbbR^omega$ such that the sequence $mathbfa - mathbfx$ is bounded. Then $mathbfa in U$.
Moreover, If $mathbfu colon= left( u_n right)_n in mathbbN in U$, then the sequence $left( a_n - u_n right)_n in mathbbN$ is bounded; that is, there exists a positive real number $r_mathbfu$ such that
$$ leftlvert a_n - u_n rightrvert < r_mathbfu mbox for all n in mathbbN. $$
So if $varepsilon in (0, 1)$ and if $mathbfx$ is any point of $mathbbR^omega$ such that $$ barrho( mathbfx, mathbfu ) < varepsilon, $$
then, for each $n in mathbbN$, we have
$$
minbig leftlvert x_n - u_n rightrvert, 1 big leq barrho( mathbfx, mathbfu ) < varepsilon < 1,
$$
and so
$$
minbig leftlvert x_n - u_n rightrvert, 1 big = leftlvert x_n - u_n rightrvert,
$$
and thus
$$ leftlvert x_n - u_n rightrvert < varepsilon. $$
Then, for any $n in mathbbN$, we obtain
$$
beginalign
leftlvert x_n - a_n rightrvert &leq leftlvert x_n - u_n rightrvert + leftlvert u_n - a_n rightrvert \
&< varepsilon + r_mathbfu.
endalign
$$
thus showing that $mathbfx in U$ also. Therefore $U$ is open in the uniform space $mathbbR^omega$.
Now the point $mathbfb$ is in the set $mathbbR^omega setminus U$.
Let $mathbfv$ be any point of $mathbbR^omega setminus U$. Then the sequence $left( v_n - a_n right)_n in mathbbN$ is unbounded; so for any natural number $k$ there exists a natural number $m_k$ such that
$$ leftlvert v_m_k - a_m_k rightrvert > k. $$
Now if $varepsilon in (0, 1)$, and if $mathbfy$ is any point of $mathbbR^omega$ such that
$$ barrho( mathbfy, mathbfv ) < varepsilon, $$
then as before we must have
$$ leftlvert y_n - v_n rightrvert < varepsilon $$
for every natural number $n$. Now if $lambda$ is an arbitrary real number and if $k$ is a natural number such that $k > lambda + varepsilon$, then we find that
$$
beginalign
leftlvert y_m_k - a_m_k rightrvert &geq leftlvert v_m_k - a_m_k rightrvert - leftlvert y_m_k - v_m_k rightrvert \
&> k - varepsilon \
&> lambda.
endalign
$$
Thus corresponding to any real number $lambda$, no matter how large, we can find a natural number $m_k$ such that
$$ leftlvert y_m_k - a_m_k rightrvert > lambda. $$
Thus the sequence $left( y_n - a_n right)_n in mathbbN$ is unbounded, which shows that $mathbfy$ is in $mathbbR^omega setminus U$. Therefore the set $mathbbR^omega setminus U$ is also open in the uniform metric space $mathbbR^omega$.
Thus the sets $U$ and $mathbbR^omega setminus U$ constitute a separation of the uniform metric space $mathbbR^omega$. Moreover, these two sets are both open and closed in $mathbbR^omega$ with the uniform topology.
Finally, if $mathbfx$ and $mathbfy$ are any two points of $U$. then the sequences $mathbfx - mathbfa$ and $mathbfy - mathbfa$ are both bounded; so the sequence $mathbfx - mathbfy = ( mathbfx - mathbfa ) - ( mathbfy - mathbfa )$ is bounded as well. Thus $mathbfx$ and $mathbfy$ are in the same (path) component of the uniform metric space $mathbbR^omega$, as has been shown above. Thus the set $U$ is (path) connected.
Similarly, the set $V$ of all the points $mathbfv$ in $mathbbR^omega$ such that $mathbfv - mathbfb$ is bounded is (path) connected. Moreover, $mathbfb$ is in $V$ but $mathbfa$ is not in $V$. Also $V$ is both open and closed in the uniform metric space $mathbbR^omega$, just as $U$ has been shown to be both open and closed.
Moreover, the sets $U$ and $V$ are disjoint.
I'm getting lost now!
What next? How to proceed from here? Or, is there anywhere I've gone wrong?
P.S.:
I think I've just managed to hit upon the right trick!
Now as the sets $U$ and $mathbbR^omega setminus U$ form a separation of the uniform metric space $mathbbR^omega$, so any connected subspace lies in either $U$ or $mathbbR^omega setminus U$, but not both. [Refer to Lemma 23.2 in Munkres.]
As $mathbfa in U$, so the component containing $mathbfa$, which is of course a connected subspace of $mathbbR^omega$, must also lie in $U$. And, as $mathbfb$ is in $mathbbR^omega setminus U$, so the component containing $mathbfb$ lies in $mathbbR^omega setminus U$. Therefore the points $mathbfa$ and $mathbfb$ lie in different components of $mathbbR^omega$ in the uniform topology if the sequence $mathbfa - mathbfb$ is unbounded.
Is there any flaw in my reasoning?
general-topology proof-verification connectedness alternative-proof path-connected
asked Jul 27 at 19:50
Saaqib Mahmood
7,11042169
edited Jul 28 at 11:42
asked Jul 27 at 19:50
Saaqib Mahmood
7,11042169
asked Jul 27 at 19:50
Saaqib Mahmood
7,11042169
asked Jul 27 at 19:50
Saaqib Mahmood
7,11042169
7,11042169
1
Does your notational distinction between $mathbf R$ and $mathbb R$ carry any significance? If so, it should be explained; if not, it should be removed.
– joriki
Jul 28 at 3:00
@joriki I write $mathbbR$ to denote the set of real numbers. I'll just edit my post to omit $mathbfR$ whereever I've used it. Thanks for the comment.
– Saaqib Mahmood
Jul 28 at 11:38
See also math.stackexchange.com/q/2730869.
– Paul Frost
Jul 28 at 13:51
add a comment |Â
1
Does your notational distinction between $mathbf R$ and $mathbb R$ carry any significance? If so, it should be explained; if not, it should be removed.
– joriki
Jul 28 at 3:00
@joriki I write $mathbbR$ to denote the set of real numbers. I'll just edit my post to omit $mathbfR$ whereever I've used it. Thanks for the comment.
– Saaqib Mahmood
Jul 28 at 11:38
See also math.stackexchange.com/q/2730869.
– Paul Frost
Jul 28 at 13:51
1
1
Does your notational distinction between $mathbf R$ and $mathbb R$ carry any significance? If so, it should be explained; if not, it should be removed.
– joriki
Jul 28 at 3:00
Does your notational distinction between $mathbf R$ and $mathbb R$ carry any significance? If so, it should be explained; if not, it should be removed.
– joriki
Jul 28 at 3:00
@joriki I write $mathbbR$ to denote the set of real numbers. I'll just edit my post to omit $mathbfR$ whereever I've used it. Thanks for the comment.
– Saaqib Mahmood
Jul 28 at 11:38
@joriki I write $mathbbR$ to denote the set of real numbers. I'll just edit my post to omit $mathbfR$ whereever I've used it. Thanks for the comment.
– Saaqib Mahmood
Jul 28 at 11:38
See also math.stackexchange.com/q/2730869.
– Paul Frost
Jul 28 at 13:51
See also math.stackexchange.com/q/2730869.
– Paul Frost
Jul 28 at 13:51
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
What you did seems great.
Some comments:
- A hint was provided in the exercise, which is that it is sufficient to consider the case where one of the sequence is always vanishing. Indeed it simplifies the notations. You can prove that considering this case is enough by considering the translation $tau_a$ such that $tau_a(mathbf 0)= mathbf a$. $tau_a$ is an homeomorphism of $mathbb R^omega$, and therefore transform components into components.
- Regarding your question to avoid using path connectedness. You can notice that the sup norm is well defined on $U$. Also, the open ball $B$ centered on $mathbf a$ with radius $2Vert mathbf a -mathbf bVert_infty$ is open, connected as all open balls and included in $U$. Therefore $U$ is connected. This is a general result that a union of connected sets that all contain a point $x$ is connected.
answered Jul 28 at 15:39
mathcounterexamples.net
23.5k21652
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
What you did seems great.
Some comments:
- A hint was provided in the exercise, which is that it is sufficient to consider the case where one of the sequence is always vanishing. Indeed it simplifies the notations. You can prove that considering this case is enough by considering the translation $tau_a$ such that $tau_a(mathbf 0)= mathbf a$. $tau_a$ is an homeomorphism of $mathbb R^omega$, and therefore transform components into components.
- Regarding your question to avoid using path connectedness. You can notice that the sup norm is well defined on $U$. Also, the open ball $B$ centered on $mathbf a$ with radius $2Vert mathbf a -mathbf bVert_infty$ is open, connected as all open balls and included in $U$. Therefore $U$ is connected. This is a general result that a union of connected sets that all contain a point $x$ is connected.
answered Jul 28 at 15:39
mathcounterexamples.net
23.5k21652
add a comment |Â
up vote
1
down vote
What you did seems great.
Some comments:
- A hint was provided in the exercise, which is that it is sufficient to consider the case where one of the sequence is always vanishing. Indeed it simplifies the notations. You can prove that considering this case is enough by considering the translation $tau_a$ such that $tau_a(mathbf 0)= mathbf a$. $tau_a$ is an homeomorphism of $mathbb R^omega$, and therefore transform components into components.
- Regarding your question to avoid using path connectedness. You can notice that the sup norm is well defined on $U$. Also, the open ball $B$ centered on $mathbf a$ with radius $2Vert mathbf a -mathbf bVert_infty$ is open, connected as all open balls and included in $U$. Therefore $U$ is connected. This is a general result that a union of connected sets that all contain a point $x$ is connected.
answered Jul 28 at 15:39
mathcounterexamples.net
23.5k21652
add a comment |Â
up vote
1
down vote
up vote
1
down vote
What you did seems great.
Some comments:
- A hint was provided in the exercise, which is that it is sufficient to consider the case where one of the sequence is always vanishing. Indeed it simplifies the notations. You can prove that considering this case is enough by considering the translation $tau_a$ such that $tau_a(mathbf 0)= mathbf a$. $tau_a$ is an homeomorphism of $mathbb R^omega$, and therefore transform components into components.
- Regarding your question to avoid using path connectedness. You can notice that the sup norm is well defined on $U$. Also, the open ball $B$ centered on $mathbf a$ with radius $2Vert mathbf a -mathbf bVert_infty$ is open, connected as all open balls and included in $U$. Therefore $U$ is connected. This is a general result that a union of connected sets that all contain a point $x$ is connected.
answered Jul 28 at 15:39
mathcounterexamples.net
23.5k21652
What you did seems great.
Some comments:
- A hint was provided in the exercise, which is that it is sufficient to consider the case where one of the sequence is always vanishing. Indeed it simplifies the notations. You can prove that considering this case is enough by considering the translation $tau_a$ such that $tau_a(mathbf 0)= mathbf a$. $tau_a$ is an homeomorphism of $mathbb R^omega$, and therefore transform components into components.
- Regarding your question to avoid using path connectedness. You can notice that the sup norm is well defined on $U$. Also, the open ball $B$ centered on $mathbf a$ with radius $2Vert mathbf a -mathbf bVert_infty$ is open, connected as all open balls and included in $U$. Therefore $U$ is connected. This is a general result that a union of connected sets that all contain a point $x$ is connected.
answered Jul 28 at 15:39
mathcounterexamples.net
23.5k21652
edited Jul 28 at 15:54
answered Jul 28 at 15:39
mathcounterexamples.net
23.5k21652
answered Jul 28 at 15:39
mathcounterexamples.net
23.5k21652
answered Jul 28 at 15:39
mathcounterexamples.net
23.5k21652
23.5k21652
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864733%2fprob-2b-sec-25-in-munkres-topology-2nd-ed-the-iff-condition-for-two-poi%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Does your notational distinction between $mathbf R$ and $mathbb R$ carry any significance? If so, it should be explained; if not, it should be removed.
– joriki
Jul 28 at 3:00
@joriki I write $mathbbR$ to denote the set of real numbers. I'll just edit my post to omit $mathbfR$ whereever I've used it. Thanks for the comment.
– Saaqib Mahmood
Jul 28 at 11:38
See also math.stackexchange.com/q/2730869.
– Paul Frost
Jul 28 at 13:51