Continuity of Function W.R.T. Topology Generated by a Function
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Let $(X,d)$ be some metric space, and let $x_0 in X$ be some fixed point. Denote by $mathcal{T}_0$ the topology with respect to which $x mapsto d(x_0,x)$ is continuous. My question is,
If $x_1 neq x_0$, will the map $x mapsto d(x_1,x)$ be continuous wrt $(X,mathcal{T}_0)$?
I've tried to prove that it is true but had no luck, so I suspect that it is false. However, I'm having trouble coming up with a counterexample. In term of trying to find a counterexample, I considered the function $f_0 : Bbb{R} to [0,infty)$ and the topology $mathcal{T}_0$ wrt it is continuous, and then tried showing that $f_1 : Bbb{R} to [0,infty)$ via $f(x) = |1-x|$ is not continuous wrt $(Bbb{R},mathcal{T}_0)$ by considering various pullbacks of closed sets. Sadly, I didn't have much luck. I could use some help sorting all of this out.
Going back to the setup at the beginning of this post, ultimately I am trying to prove that $mathcal{T}_0$ is finer than the metric topology. If I could show that $x mapsto d(x_1,x)$ is in fact continuous wrt $(X,mathcal{T}_0)$, then I would have an easy proof that $mathcal{T}_0$ is finer than the metric topology.
general-topology continuity metric-spaces
asked Dec 19 '18 at 19:56
user193319user193319
2,4212926
$endgroup$
add a comment |
$begingroup$
Let $(X,d)$ be some metric space, and let $x_0 in X$ be some fixed point. Denote by $mathcal{T}_0$ the topology with respect to which $x mapsto d(x_0,x)$ is continuous. My question is,
If $x_1 neq x_0$, will the map $x mapsto d(x_1,x)$ be continuous wrt $(X,mathcal{T}_0)$?
I've tried to prove that it is true but had no luck, so I suspect that it is false. However, I'm having trouble coming up with a counterexample. In term of trying to find a counterexample, I considered the function $f_0 : Bbb{R} to [0,infty)$ and the topology $mathcal{T}_0$ wrt it is continuous, and then tried showing that $f_1 : Bbb{R} to [0,infty)$ via $f(x) = |1-x|$ is not continuous wrt $(Bbb{R},mathcal{T}_0)$ by considering various pullbacks of closed sets. Sadly, I didn't have much luck. I could use some help sorting all of this out.
Going back to the setup at the beginning of this post, ultimately I am trying to prove that $mathcal{T}_0$ is finer than the metric topology. If I could show that $x mapsto d(x_1,x)$ is in fact continuous wrt $(X,mathcal{T}_0)$, then I would have an easy proof that $mathcal{T}_0$ is finer than the metric topology.
general-topology continuity metric-spaces
asked Dec 19 '18 at 19:56
user193319user193319
2,4212926
$endgroup$
$begingroup$
I think either the old prelim has an error or you have copied it wrong. First of all, there is no such thing as "the topology with respect to which $xmapsto d(x_0,x)$". Second of all, assuming $mathcal{T}_0$ is just supposed to be an arbitrary topology with this property, you need the property to hold for every $x_0in X$, not just for some fixed $x_0$. It is not true if you just have one $x_0$.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:33
$begingroup$
@EricWofsey You're right. The problem uses the indefinite article, but I am not sure I follow your second point. Here's the prelim: albany.edu/math/images/prelims/top1601.pdf
$endgroup$
– user193319
Dec 19 '18 at 20:37
$begingroup$
Yeah, the problem definitely intends to have a universal quantifier on $x_0$, not an existential quantifier. The phrasing with "fix a point $x_0in X$" at the start is a mistake. (It is correct for the first half of the problem but not the second half.)
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:37
$begingroup$
@EricWofsey Ah, so $mathcal{T}_0$ is a topology on $X$ with respect to which $y mapsto d(x,y)$ is continuous for every $x in X$?
$endgroup$
– user193319
Dec 19 '18 at 20:41
2
$begingroup$
Right, that's the intended meaning of the second part.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:42
add a comment |
$begingroup$
Let $(X,d)$ be some metric space, and let $x_0 in X$ be some fixed point. Denote by $mathcal{T}_0$ the topology with respect to which $x mapsto d(x_0,x)$ is continuous. My question is,
If $x_1 neq x_0$, will the map $x mapsto d(x_1,x)$ be continuous wrt $(X,mathcal{T}_0)$?
I've tried to prove that it is true but had no luck, so I suspect that it is false. However, I'm having trouble coming up with a counterexample. In term of trying to find a counterexample, I considered the function $f_0 : Bbb{R} to [0,infty)$ and the topology $mathcal{T}_0$ wrt it is continuous, and then tried showing that $f_1 : Bbb{R} to [0,infty)$ via $f(x) = |1-x|$ is not continuous wrt $(Bbb{R},mathcal{T}_0)$ by considering various pullbacks of closed sets. Sadly, I didn't have much luck. I could use some help sorting all of this out.
Going back to the setup at the beginning of this post, ultimately I am trying to prove that $mathcal{T}_0$ is finer than the metric topology. If I could show that $x mapsto d(x_1,x)$ is in fact continuous wrt $(X,mathcal{T}_0)$, then I would have an easy proof that $mathcal{T}_0$ is finer than the metric topology.
general-topology continuity metric-spaces
asked Dec 19 '18 at 19:56
user193319user193319
2,4212926
$endgroup$
Let $(X,d)$ be some metric space, and let $x_0 in X$ be some fixed point. Denote by $mathcal{T}_0$ the topology with respect to which $x mapsto d(x_0,x)$ is continuous. My question is,
If $x_1 neq x_0$, will the map $x mapsto d(x_1,x)$ be continuous wrt $(X,mathcal{T}_0)$?
I've tried to prove that it is true but had no luck, so I suspect that it is false. However, I'm having trouble coming up with a counterexample. In term of trying to find a counterexample, I considered the function $f_0 : Bbb{R} to [0,infty)$ and the topology $mathcal{T}_0$ wrt it is continuous, and then tried showing that $f_1 : Bbb{R} to [0,infty)$ via $f(x) = |1-x|$ is not continuous wrt $(Bbb{R},mathcal{T}_0)$ by considering various pullbacks of closed sets. Sadly, I didn't have much luck. I could use some help sorting all of this out.
Going back to the setup at the beginning of this post, ultimately I am trying to prove that $mathcal{T}_0$ is finer than the metric topology. If I could show that $x mapsto d(x_1,x)$ is in fact continuous wrt $(X,mathcal{T}_0)$, then I would have an easy proof that $mathcal{T}_0$ is finer than the metric topology.
general-topology continuity metric-spaces
general-topology continuity metric-spaces
asked Dec 19 '18 at 19:56
user193319user193319
2,4212926
asked Dec 19 '18 at 19:56
user193319user193319
2,4212926
edited Dec 19 '18 at 20:19
user193319
asked Dec 19 '18 at 19:56
user193319user193319
2,4212926
asked Dec 19 '18 at 19:56
user193319user193319
2,4212926
asked Dec 19 '18 at 19:56
user193319user193319
2,4212926
2,4212926
$begingroup$
I think either the old prelim has an error or you have copied it wrong. First of all, there is no such thing as "the topology with respect to which $xmapsto d(x_0,x)$". Second of all, assuming $mathcal{T}_0$ is just supposed to be an arbitrary topology with this property, you need the property to hold for every $x_0in X$, not just for some fixed $x_0$. It is not true if you just have one $x_0$.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:33
$begingroup$
@EricWofsey You're right. The problem uses the indefinite article, but I am not sure I follow your second point. Here's the prelim: albany.edu/math/images/prelims/top1601.pdf
$endgroup$
– user193319
Dec 19 '18 at 20:37
$begingroup$
Yeah, the problem definitely intends to have a universal quantifier on $x_0$, not an existential quantifier. The phrasing with "fix a point $x_0in X$" at the start is a mistake. (It is correct for the first half of the problem but not the second half.)
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:37
$begingroup$
@EricWofsey Ah, so $mathcal{T}_0$ is a topology on $X$ with respect to which $y mapsto d(x,y)$ is continuous for every $x in X$?
$endgroup$
– user193319
Dec 19 '18 at 20:41
2
$begingroup$
Right, that's the intended meaning of the second part.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:42
add a comment |
$begingroup$
I think either the old prelim has an error or you have copied it wrong. First of all, there is no such thing as "the topology with respect to which $xmapsto d(x_0,x)$". Second of all, assuming $mathcal{T}_0$ is just supposed to be an arbitrary topology with this property, you need the property to hold for every $x_0in X$, not just for some fixed $x_0$. It is not true if you just have one $x_0$.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:33
$begingroup$
@EricWofsey You're right. The problem uses the indefinite article, but I am not sure I follow your second point. Here's the prelim: albany.edu/math/images/prelims/top1601.pdf
$endgroup$
– user193319
Dec 19 '18 at 20:37
$begingroup$
Yeah, the problem definitely intends to have a universal quantifier on $x_0$, not an existential quantifier. The phrasing with "fix a point $x_0in X$" at the start is a mistake. (It is correct for the first half of the problem but not the second half.)
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:37
$begingroup$
@EricWofsey Ah, so $mathcal{T}_0$ is a topology on $X$ with respect to which $y mapsto d(x,y)$ is continuous for every $x in X$?
$endgroup$
– user193319
Dec 19 '18 at 20:41
2
$begingroup$
Right, that's the intended meaning of the second part.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:42
$begingroup$
I think either the old prelim has an error or you have copied it wrong. First of all, there is no such thing as "the topology with respect to which $xmapsto d(x_0,x)$". Second of all, assuming $mathcal{T}_0$ is just supposed to be an arbitrary topology with this property, you need the property to hold for every $x_0in X$, not just for some fixed $x_0$. It is not true if you just have one $x_0$.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:33
$begingroup$
I think either the old prelim has an error or you have copied it wrong. First of all, there is no such thing as "the topology with respect to which $xmapsto d(x_0,x)$". Second of all, assuming $mathcal{T}_0$ is just supposed to be an arbitrary topology with this property, you need the property to hold for every $x_0in X$, not just for some fixed $x_0$. It is not true if you just have one $x_0$.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:33
$begingroup$
@EricWofsey You're right. The problem uses the indefinite article, but I am not sure I follow your second point. Here's the prelim: albany.edu/math/images/prelims/top1601.pdf
$endgroup$
– user193319
Dec 19 '18 at 20:37
$begingroup$
@EricWofsey You're right. The problem uses the indefinite article, but I am not sure I follow your second point. Here's the prelim: albany.edu/math/images/prelims/top1601.pdf
$endgroup$
– user193319
Dec 19 '18 at 20:37
$begingroup$
Yeah, the problem definitely intends to have a universal quantifier on $x_0$, not an existential quantifier. The phrasing with "fix a point $x_0in X$" at the start is a mistake. (It is correct for the first half of the problem but not the second half.)
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:37
$begingroup$
Yeah, the problem definitely intends to have a universal quantifier on $x_0$, not an existential quantifier. The phrasing with "fix a point $x_0in X$" at the start is a mistake. (It is correct for the first half of the problem but not the second half.)
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:37
$begingroup$
@EricWofsey Ah, so $mathcal{T}_0$ is a topology on $X$ with respect to which $y mapsto d(x,y)$ is continuous for every $x in X$?
$endgroup$
– user193319
Dec 19 '18 at 20:41
$begingroup$
@EricWofsey Ah, so $mathcal{T}_0$ is a topology on $X$ with respect to which $y mapsto d(x,y)$ is continuous for every $x in X$?
$endgroup$
– user193319
Dec 19 '18 at 20:41
2
2
$begingroup$
Right, that's the intended meaning of the second part.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:42
$begingroup$
Right, that's the intended meaning of the second part.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:42
add a comment |
1 Answer
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Here is an answer if you read the question verbatim: consider $mathbb R$ with the usual metric. One topology which makes $x to d(x,0)$ continuous is the collection of sets of the form ${x:|x| in U}$ where $U$ is open in $mathbb R$. The function $x to d(x,1)$ is not continuous in this topology because ${x: |x-1| <1}=(0,2)$ is not of the form ${x:|x| in U}$ for any open set $U$. [ Sets of the the latter type are all symmetric about $0$].
answered Dec 19 '18 at 23:44
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
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add a comment |
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$begingroup$
Here is an answer if you read the question verbatim: consider $mathbb R$ with the usual metric. One topology which makes $x to d(x,0)$ continuous is the collection of sets of the form ${x:|x| in U}$ where $U$ is open in $mathbb R$. The function $x to d(x,1)$ is not continuous in this topology because ${x: |x-1| <1}=(0,2)$ is not of the form ${x:|x| in U}$ for any open set $U$. [ Sets of the the latter type are all symmetric about $0$].
answered Dec 19 '18 at 23:44
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
$endgroup$
add a comment |
$begingroup$
Here is an answer if you read the question verbatim: consider $mathbb R$ with the usual metric. One topology which makes $x to d(x,0)$ continuous is the collection of sets of the form ${x:|x| in U}$ where $U$ is open in $mathbb R$. The function $x to d(x,1)$ is not continuous in this topology because ${x: |x-1| <1}=(0,2)$ is not of the form ${x:|x| in U}$ for any open set $U$. [ Sets of the the latter type are all symmetric about $0$].
answered Dec 19 '18 at 23:44
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
$endgroup$
add a comment |
$begingroup$
Here is an answer if you read the question verbatim: consider $mathbb R$ with the usual metric. One topology which makes $x to d(x,0)$ continuous is the collection of sets of the form ${x:|x| in U}$ where $U$ is open in $mathbb R$. The function $x to d(x,1)$ is not continuous in this topology because ${x: |x-1| <1}=(0,2)$ is not of the form ${x:|x| in U}$ for any open set $U$. [ Sets of the the latter type are all symmetric about $0$].
answered Dec 19 '18 at 23:44
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
$endgroup$
Here is an answer if you read the question verbatim: consider $mathbb R$ with the usual metric. One topology which makes $x to d(x,0)$ continuous is the collection of sets of the form ${x:|x| in U}$ where $U$ is open in $mathbb R$. The function $x to d(x,1)$ is not continuous in this topology because ${x: |x-1| <1}=(0,2)$ is not of the form ${x:|x| in U}$ for any open set $U$. [ Sets of the the latter type are all symmetric about $0$].
answered Dec 19 '18 at 23:44
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
answered Dec 19 '18 at 23:44
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
answered Dec 19 '18 at 23:44
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
answered Dec 19 '18 at 23:44
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
63.9k42464
add a comment |
add a comment |
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$begingroup$
I think either the old prelim has an error or you have copied it wrong. First of all, there is no such thing as "the topology with respect to which $xmapsto d(x_0,x)$". Second of all, assuming $mathcal{T}_0$ is just supposed to be an arbitrary topology with this property, you need the property to hold for every $x_0in X$, not just for some fixed $x_0$. It is not true if you just have one $x_0$.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:33
$begingroup$
@EricWofsey You're right. The problem uses the indefinite article, but I am not sure I follow your second point. Here's the prelim: albany.edu/math/images/prelims/top1601.pdf
$endgroup$
– user193319
Dec 19 '18 at 20:37
$begingroup$
Yeah, the problem definitely intends to have a universal quantifier on $x_0$, not an existential quantifier. The phrasing with "fix a point $x_0in X$" at the start is a mistake. (It is correct for the first half of the problem but not the second half.)
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:37
$begingroup$
@EricWofsey Ah, so $mathcal{T}_0$ is a topology on $X$ with respect to which $y mapsto d(x,y)$ is continuous for every $x in X$?
$endgroup$
– user193319
Dec 19 '18 at 20:41
2
$begingroup$
Right, that's the intended meaning of the second part.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 20:42