Return to Answer

If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk and $x(t)$ passes through the boundary of the disk and later intersects the disk).

To begin with, let $t_0=\tau^V(x_0)$. We can give local coordinates $(x^1,\dots,x^n)$ for some neighborhood $U$ of $x(t_0)$ so that $V\cap U=\{x^1=0\}\subseteq U$. Now as $v(x(t_0))\not\in T_x(v)$, we can suppose that, for some $\varepsilon>0$, $x^1(x(t))<0$ for all $t\in(t_0-\varepsilon,t_0)$ and $x^1(x(t))>0$ for $t\in(t_0,t_0+\varepsilon)$. As the flow is continuous, there is a neighborhood $V\times(t_0-\delta,t_0+\delta)$ of $(x_0,t_0)\in\mathcal{O}\times\mathbb{R}$ (with $\delta<\varepsilon$) such that $y(t)\in U$ for all $(y,t)\in V\times(t_0-\delta,t_0+\delta)$ (where $y(t)$ is defined by $y(0)=y,y'=v(y)$). Moreover, as $x^1(x(t_0+\delta))>0$ and $x^1(x(t_0-\delta))<0$, shrinking the neighborhood $V$ if necessary we get that $x^1(y(t_0+\delta))>0$ and $x^1(y(t_0-\delta))<0$ for all $y\in V$. So by the intermediate value theorem, for every $y\in V$ there is some $s\in(t_0-\delta,t_0+\delta)$ such that $y(s)\in V$.

As we can make $\delta$ as small as we want, we get that for all $\delta>0$ there is a neighborhood $V$ of $x$ such that $\tau^V(y)<t_0+\delta\;\forall y\in V$. (We have still not used that $V$ is closed)

Now we just need to prove that for all $\delta>0$ there is a neighborhood $V$ of $x$ such that $\tau^V(y)>t_0-\delta\;\forall y\in V$. But if this was false for some $\delta$, there would be a sequence of points $p_n\to x$ and a sequence of times $t_n<t_0-\delta$ such that $p_n(t_n)\in V$. As $V$ is closed (and once again using continuity of the flow), this means that for some accumulation point $t\leq t_0-\delta$ of the sequence $t_n$ we have $x(t)\in V$, a contradiction.

If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk and $x(t)$ passes through the boundary of the disk and later intersects the disk).

To begin with, let $t_0=\tau^V(x_0)$. We can give local coordinates $(x^1,\dots,x^n)$ for some neighborhood $U$ of $x(t_0)$ so that $V\cap U=\{x^1=0\}\subseteq U$. Now as $v(x(t_0))\not\in T_x(v)$, we can suppose that, for some $\varepsilon>0$, $x^1(x(t))<0$ for all $t\in(t_0-\varepsilon,t_0)$ and $x^1(x(t))>0$ for $t\in(t_0,t_0+\varepsilon)$. As the flow is continuous, there is a neighborhood $W\times(t_0-\delta,t_0+\delta)$ of $(x_0,t_0)\in\mathcal{O}\times\mathbb{R}$ (with $\delta<\varepsilon$) such that $y(t)\in U$ for all $(y,t)\in W\times(t_0-\delta,t_0+\delta)$ (where $y(t)$ is defined by $y(0)=y,y'=v(y)$). Moreover, as $x^1(x(t_0+\delta))>0$ and $x^1(x(t_0-\delta))<0$, shrinking the neighborhood $W$ if necessary we get that $x^1(y(t_0+\delta))>0$ and $x^1(y(t_0-\delta))<0$ for all $y\in W$. So by the intermediate value theorem, for every $y\in W$ there is some $s\in(t_0-\delta,t_0+\delta)$ such that $y(s)\in W$.

As we can make $\delta$ as small as we want, we get that for all $\delta>0$ there is a neighborhood $W$ of $x$ such that $\tau^V(y)<t_0+\delta\;\forall y\in W$. (We have still not used that $V$ is closed)

Now we just need to prove that for all $\delta>0$ there is a neighborhood $W$ of $x$ such that $\tau^V(y)>t_0-\delta\;\forall y\in W$. But if this was false for some $\delta$, there would be a sequence of points $p_n\to x$ and a sequence of times $t_n<t_0-\delta$ such that $p_n(t_n)\in V$. As $V$ is closed (and once again using continuity of the flow), this means that for some accumulation point $t\leq t_0-\delta$ of the sequence $t_n$ we have $x(t)\in V$, a contradiction.

If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk and $x(t)$ passes through the boundary of the disk and later intersects the disk).

To begin with, let $t_0=\tau^V(x_0)$. We can give local coordinates $(x^1,\dots,x^n)$ for some neighborhood $U$ of $x(t_0)$ so that $V\cap U=\{x^1=0\}\subseteq U$. Now as $v(x(t_0))\not\in T_x(v)$, we can suppose that, for some $\varepsilon>0$, $x^1(x(t))<0$ for all $t\in(t_0-\varepsilon,t_0)$ and $x^1(x(t))>0$ for $t\in(t_0,t_0+\varepsilon)$. As the flow is continuous, there is a neighborhood $V\times(t_0-\delta,t_0+\delta)$ of $(x_0,t_0)\in\mathcal{O}\times\mathbb{R}$ (with $\delta<\varepsilon$) such that $y(t)\in U$ for all $(y,t)\in V\times(t_0-\delta,t_0+\delta)$ (where $y(t)$ is defined by $y(0)=y,y'=v(y)$). Moreover, as $x^1(x(t_0+\delta))>0$ and $x^1(x(t_0-\delta))<0$, shrinking the neighborhood $V$ if necessary we get that $x^1(y(t_0+\delta))>0$ and $x^1(y(t_0-\delta))<0$ for all $y\in V$. So by the intermediate value theorem, for every $y\in V$ there is some $s\in(t_0-\delta,t_0+\delta)$ such that $y(s)\in V$.

As we can make $\delta$ as small as we want, we get that for all $\delta>0$ there is a neighborhood $V$ of $x$ such that $\tau^V(y)<t_0+\delta\;\forall y\in V$. (We have still not used that $V$ is closed)

Now we just need to prove that for all $\delta>0$ there is a neighborhood $V$ of $x$ such that $\tau^V(y)>t_0-\delta\;\forall y\in V$. But if this was false for some $\delta$, there would be a sequence of points $p_n\to x$ and a sequence of times $t_n<t_0-\delta$ such that $p_n(t_n)\in V$. As $V$ is closed (and once again using continuity of the flow), this means that for some accumulation point $t\leq\tau^V(x_0)-\delta$ of the sequence $t_n$ we have $x(t)\in V$, a contradiction.

If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk and $x(t)$ passes through the boundary of the disk and later intersects the disk).

To begin with, let $t_0=\tau^V(x_0)$. We can give local coordinates $(x^1,\dots,x^n)$ for some neighborhood $U$ of $x(t_0)$ so that $V\cap U=\{x^1=0\}\subseteq U$. Now as $v(x(t_0))\not\in T_x(v)$, we can suppose that, for some $\varepsilon>0$, $x^1(x(t))<0$ for all $t\in(t_0-\varepsilon,t_0)$ and $x^1(x(t))>0$ for $t\in(t_0,t_0+\varepsilon)$. As the flow is continuous, there is a neighborhood $V\times(t_0-\delta,t_0+\delta)$ of $(x_0,t_0)\in\mathcal{O}\times\mathbb{R}$ (with $\delta<\varepsilon$) such that $y(t)\in U$ for all $(y,t)\in V\times(t_0-\delta,t_0+\delta)$ (where $y(t)$ is defined by $y(0)=y,y'=v(y)$). Moreover, as $x^1(x(t_0+\delta))>0$ and $x^1(x(t_0-\delta))<0$, shrinking the neighborhood $V$ if necessary we get that $x^1(y(t_0+\delta))>0$ and $x^1(y(t_0-\delta))<0$ for all $y\in V$. So by the intermediate value theorem, for every $y\in V$ there is some $s\in(t_0-\delta,t_0+\delta)$ such that $y(s)\in V$.

As we can make $\delta$ as small as we want, we get that for all $\delta>0$ there is a neighborhood $V$ of $x$ such that $\tau^V(y)<t_0+\delta\;\forall y\in V$. (We have still not used that $V$ is closed)

Now we just need to prove that for all $\delta>0$ there is a neighborhood $V$ of $x$ such that $\tau^V(y)>t_0-\delta\;\forall y\in V$. But if this was false for some $\delta$, there would be a sequence of points $p_n\to x$ and a sequence of times $t_n<t_0-\delta$ such that $p_n(t_n)\in V$. As $V$ is closed (and once again using continuity of the flow), this means that for some accumulation point $t\leq t_0-\delta$ of the sequence $t_n$ we have $x(t)\in V$, a contradiction.

If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk and $x(t)$ passes through the boundary of the disk and later intersects the disk).

To begin with, let $t_0=\tau^V(x_0)$. We can give local coordinates $(x^1,\dots,x^n)$ for some neighborhood $U$ of $x(t_0)$ so that $V\cap U=\{x^1=0\}\subseteq U$. Now as $v(x(t_0))\not\in T_x(v)$, we can suppose that, for some $\varepsilon>0$, $x^1(x(t))<0$ for all $t\in(t_0-\varepsilon,t_0)$ and $x^1(x(t))>0$ for $t\in(t_0,t_0+\varepsilon)$. As the flow is continuous, there is a neighborhood $V\times(t-\delta,t+\delta)$ of $(x_0,t)\in\mathcal{O}\times\mathbb{R}$ (with $\delta<\varepsilon$) such that $y(t)\in U$ for all $(y,t)\in V\times(t-\delta,t+\delta)$ (where $y(t)$ is defined by $y(0)=y,y'=v(y)$). Moreover, as $x^1(x(t+\delta))>0$ and $x^1(x(t-\delta))<0$, shrinking the neighborhood $V$ if necessary we get that $x^1(y(t+\delta))>0$ and $x^1(y(t-\delta))<0$ for all $y\in V$. So by the intermediate value theorem, for every $y\in V$ there is some $s\in(t-\delta,t+\delta)$ such that $y(s)\in V$.

As we can make $\delta$ as small as we want, we get that for all $\delta>0$ there is a neighborhood $V$ of $x$ such that $\tau^V(y)<\tau^V(x_0)+\delta\;\forall y\in V$. (We have still not used that $V$ is closed)

Now we just need to prove that for all $\delta>0$ there is a neighborhood $V$ of $x$ such that $\tau^V(y)>\tau^V(x_0)-\delta\;\forall y\in V$. But if this was false for some $\delta$, there would be a sequence of points $p_n\to x$ and a sequence of times $t_n<\tau^V(x_0)-\delta$ such that $p_n(t_n)\in V$. As $V$ is closed (and once again using continuity of the flow), this means that for some accumulation point $t\leq\tau^V(x_0)-\delta$ of the sequence $t_n$ we have $x(t)\in V$, a contradiction.

If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk and $x(t)$ passes through the boundary of the disk and later intersects the disk).

To begin with, let $t_0=\tau^V(x_0)$. We can give local coordinates $(x^1,\dots,x^n)$ for some neighborhood $U$ of $x(t_0)$ so that $V\cap U=\{x^1=0\}\subseteq U$. Now as $v(x(t_0))\not\in T_x(v)$, we can suppose that, for some $\varepsilon>0$, $x^1(x(t))<0$ for all $t\in(t_0-\varepsilon,t_0)$ and $x^1(x(t))>0$ for $t\in(t_0,t_0+\varepsilon)$. As the flow is continuous, there is a neighborhood $V\times(t_0-\delta,t_0+\delta)$ of $(x_0,t_0)\in\mathcal{O}\times\mathbb{R}$ (with $\delta<\varepsilon$) such that $y(t)\in U$ for all $(y,t)\in V\times(t_0-\delta,t_0+\delta)$ (where $y(t)$ is defined by $y(0)=y,y'=v(y)$). Moreover, as $x^1(x(t_0+\delta))>0$ and $x^1(x(t_0-\delta))<0$, shrinking the neighborhood $V$ if necessary we get that $x^1(y(t_0+\delta))>0$ and $x^1(y(t_0-\delta))<0$ for all $y\in V$. So by the intermediate value theorem, for every $y\in V$ there is some $s\in(t_0-\delta,t_0+\delta)$ such that $y(s)\in V$.

As we can make $\delta$ as small as we want, we get that for all $\delta>0$ there is a neighborhood $V$ of $x$ such that $\tau^V(y)<t_0+\delta\;\forall y\in V$. (We have still not used that $V$ is closed)

Now we just need to prove that for all $\delta>0$ there is a neighborhood $V$ of $x$ such that $\tau^V(y)>t_0-\delta\;\forall y\in V$. But if this was false for some $\delta$, there would be a sequence of points $p_n\to x$ and a sequence of times $t_n<t_0-\delta$ such that $p_n(t_n)\in V$. As $V$ is closed (and once again using continuity of the flow), this means that for some accumulation point $t\leq\tau^V(x_0)-\delta$ of the sequence $t_n$ we have $x(t)\in V$, a contradiction.