Non-void intersection of two sets
Let $(t,x)$ be in $(0,infty)times(0,1$) with the constraint $x+t<1$. Let $(a,b)$ be a subset of $(0,1)$. I want to know for which values of $t$ we have $${(x,x+t)}cap{(a,b)}neqemptyset$$
or $${(x-t,x)}cap{(a,b)}neqemptyset$$ with the contraint $x-t<1$.
Thank you.
real-analysis analysis optimization
asked Nov 23 '18 at 13:42
GustaveGustave
720211
add a comment |
Let $(t,x)$ be in $(0,infty)times(0,1$) with the constraint $x+t<1$. Let $(a,b)$ be a subset of $(0,1)$. I want to know for which values of $t$ we have $${(x,x+t)}cap{(a,b)}neqemptyset$$
or $${(x-t,x)}cap{(a,b)}neqemptyset$$ with the contraint $x-t<1$.
Thank you.
real-analysis analysis optimization
asked Nov 23 '18 at 13:42
GustaveGustave
720211
add a comment |
Let $(t,x)$ be in $(0,infty)times(0,1$) with the constraint $x+t<1$. Let $(a,b)$ be a subset of $(0,1)$. I want to know for which values of $t$ we have $${(x,x+t)}cap{(a,b)}neqemptyset$$
or $${(x-t,x)}cap{(a,b)}neqemptyset$$ with the contraint $x-t<1$.
Thank you.
real-analysis analysis optimization
asked Nov 23 '18 at 13:42
GustaveGustave
720211
Let $(t,x)$ be in $(0,infty)times(0,1$) with the constraint $x+t<1$. Let $(a,b)$ be a subset of $(0,1)$. I want to know for which values of $t$ we have $${(x,x+t)}cap{(a,b)}neqemptyset$$
or $${(x-t,x)}cap{(a,b)}neqemptyset$$ with the contraint $x-t<1$.
Thank you.
real-analysis analysis optimization
real-analysis analysis optimization
asked Nov 23 '18 at 13:42
GustaveGustave
720211
asked Nov 23 '18 at 13:42
GustaveGustave
720211
edited Nov 23 '18 at 14:32
Gustave
asked Nov 23 '18 at 13:42
GustaveGustave
720211
asked Nov 23 '18 at 13:42
GustaveGustave
720211
asked Nov 23 '18 at 13:42
GustaveGustave
720211
720211
add a comment |
add a comment |
1 Answer
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Its strange that you let $t in (0, infty)$ and $x in (0,1)$ but then restrict $t+x<1$. This means $t$ can't possibly be more than $1$. Also the restriction $x-t<1$ holds for all elements of this space since $t$ is positive and $x$ is less than $1$.
That said, if $x>b$ then $(x,x+t) cap (a,b) = emptyset$. If $x in [a,b)$ then $(x,x+t) cap (a,b)$ is never empty. Finally if $x<a$ then $(x,x+t) cap (a,b)$ is non empty if and only if $x+t>a$.
The $(x-t,x)$ case is similar.
answered Nov 23 '18 at 14:44
EricEric
2068
Thanks @Eric...
– Gustave
Nov 23 '18 at 15:14
add a comment |
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1 Answer
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1 Answer
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Its strange that you let $t in (0, infty)$ and $x in (0,1)$ but then restrict $t+x<1$. This means $t$ can't possibly be more than $1$. Also the restriction $x-t<1$ holds for all elements of this space since $t$ is positive and $x$ is less than $1$.
That said, if $x>b$ then $(x,x+t) cap (a,b) = emptyset$. If $x in [a,b)$ then $(x,x+t) cap (a,b)$ is never empty. Finally if $x<a$ then $(x,x+t) cap (a,b)$ is non empty if and only if $x+t>a$.
The $(x-t,x)$ case is similar.
answered Nov 23 '18 at 14:44
EricEric
2068
Thanks @Eric...
– Gustave
Nov 23 '18 at 15:14
add a comment |
Its strange that you let $t in (0, infty)$ and $x in (0,1)$ but then restrict $t+x<1$. This means $t$ can't possibly be more than $1$. Also the restriction $x-t<1$ holds for all elements of this space since $t$ is positive and $x$ is less than $1$.
That said, if $x>b$ then $(x,x+t) cap (a,b) = emptyset$. If $x in [a,b)$ then $(x,x+t) cap (a,b)$ is never empty. Finally if $x<a$ then $(x,x+t) cap (a,b)$ is non empty if and only if $x+t>a$.
The $(x-t,x)$ case is similar.
answered Nov 23 '18 at 14:44
EricEric
2068
Thanks @Eric...
– Gustave
Nov 23 '18 at 15:14
add a comment |
Its strange that you let $t in (0, infty)$ and $x in (0,1)$ but then restrict $t+x<1$. This means $t$ can't possibly be more than $1$. Also the restriction $x-t<1$ holds for all elements of this space since $t$ is positive and $x$ is less than $1$.
That said, if $x>b$ then $(x,x+t) cap (a,b) = emptyset$. If $x in [a,b)$ then $(x,x+t) cap (a,b)$ is never empty. Finally if $x<a$ then $(x,x+t) cap (a,b)$ is non empty if and only if $x+t>a$.
The $(x-t,x)$ case is similar.
answered Nov 23 '18 at 14:44
EricEric
2068
Its strange that you let $t in (0, infty)$ and $x in (0,1)$ but then restrict $t+x<1$. This means $t$ can't possibly be more than $1$. Also the restriction $x-t<1$ holds for all elements of this space since $t$ is positive and $x$ is less than $1$.
That said, if $x>b$ then $(x,x+t) cap (a,b) = emptyset$. If $x in [a,b)$ then $(x,x+t) cap (a,b)$ is never empty. Finally if $x<a$ then $(x,x+t) cap (a,b)$ is non empty if and only if $x+t>a$.
The $(x-t,x)$ case is similar.
answered Nov 23 '18 at 14:44
EricEric
2068
answered Nov 23 '18 at 14:44
EricEric
2068
answered Nov 23 '18 at 14:44
EricEric
2068
answered Nov 23 '18 at 14:44
EricEric
2068
2068
Thanks @Eric...
– Gustave
Nov 23 '18 at 15:14
add a comment |
Thanks @Eric...
– Gustave
Nov 23 '18 at 15:14
Thanks @Eric...
– Gustave
Nov 23 '18 at 15:14
Thanks @Eric...
– Gustave
Nov 23 '18 at 15:14
add a comment |
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