Continuous function on a metric space/topology
$begingroup$
X a metric space. Let $gamma$1, $gamma$2 : [0,1] → X be two continuous paths such that $gamma$1 (1) = $gamma$2 (0). Consider the map$gamma$ : [0,1] → X defined as $gamma$(t) = $gamma$1 (2t), if t∈ [0,$frac{1}{2}$)
or $gamma$(t) = $gamma$2 (2t−1), if t ∈ [$frac{1}{2}$,1] Show that $gamma$ is a continuous path in X.
How should I show if this is continuous?
Can I use the "usual" definitions, as the epsilon-delta or the limit definition? Or do I have to use something else since X is a metric space?
Any guidence or help would be greatly appreciated!
functions continuity
asked Jan 5 at 13:42
quisquis
83
$endgroup$
add a comment |
$begingroup$
X a metric space. Let $gamma$1, $gamma$2 : [0,1] → X be two continuous paths such that $gamma$1 (1) = $gamma$2 (0). Consider the map$gamma$ : [0,1] → X defined as $gamma$(t) = $gamma$1 (2t), if t∈ [0,$frac{1}{2}$)
or $gamma$(t) = $gamma$2 (2t−1), if t ∈ [$frac{1}{2}$,1] Show that $gamma$ is a continuous path in X.
How should I show if this is continuous?
Can I use the "usual" definitions, as the epsilon-delta or the limit definition? Or do I have to use something else since X is a metric space?
Any guidence or help would be greatly appreciated!
functions continuity
asked Jan 5 at 13:42
quisquis
83
$endgroup$
$begingroup$
Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
$endgroup$
– Yadati Kiran
Jan 5 at 14:07
add a comment |
$begingroup$
X a metric space. Let $gamma$1, $gamma$2 : [0,1] → X be two continuous paths such that $gamma$1 (1) = $gamma$2 (0). Consider the map$gamma$ : [0,1] → X defined as $gamma$(t) = $gamma$1 (2t), if t∈ [0,$frac{1}{2}$)
or $gamma$(t) = $gamma$2 (2t−1), if t ∈ [$frac{1}{2}$,1] Show that $gamma$ is a continuous path in X.
How should I show if this is continuous?
Can I use the "usual" definitions, as the epsilon-delta or the limit definition? Or do I have to use something else since X is a metric space?
Any guidence or help would be greatly appreciated!
functions continuity
asked Jan 5 at 13:42
quisquis
83
$endgroup$
X a metric space. Let $gamma$1, $gamma$2 : [0,1] → X be two continuous paths such that $gamma$1 (1) = $gamma$2 (0). Consider the map$gamma$ : [0,1] → X defined as $gamma$(t) = $gamma$1 (2t), if t∈ [0,$frac{1}{2}$)
or $gamma$(t) = $gamma$2 (2t−1), if t ∈ [$frac{1}{2}$,1] Show that $gamma$ is a continuous path in X.
How should I show if this is continuous?
Can I use the "usual" definitions, as the epsilon-delta or the limit definition? Or do I have to use something else since X is a metric space?
Any guidence or help would be greatly appreciated!
functions continuity
functions continuity
asked Jan 5 at 13:42
quisquis
83
asked Jan 5 at 13:42
quisquis
83
asked Jan 5 at 13:42
quisquis
83
asked Jan 5 at 13:42
quisquis
83
asked Jan 5 at 13:42
quisquis
83
83
$begingroup$
Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
$endgroup$
– Yadati Kiran
Jan 5 at 14:07
add a comment |
$begingroup$
Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
$endgroup$
– Yadati Kiran
Jan 5 at 14:07
$begingroup$
Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
$endgroup$
– Yadati Kiran
Jan 5 at 14:07
$begingroup$
Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
$endgroup$
– Yadati Kiran
Jan 5 at 14:07
add a comment |
1 Answer
1
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oldest
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$begingroup$
It is also true more generally if $X$ is a topological space not necessarily metrizable.
In general if $A,B$ are closed subsets of topological space $Y$ and we have function $f:Ato X$ continuous on $A$, function $g:Bto X$ continuous on $B$ such that $f$ and $g$ coincide on $Acap B$ then the function $h=fcup g:Acup Bto X$ is continuous as well.
To prove that $h$ is continuous it is enough to show that the preimage of a closed set $Fsubseteq X$ is a closed subset of $Y$.
This preimage is the set $f^{-1}(F)cup g^{-1}(F)$, so we are ready if we can prove that the sets $f^{-1}(F)$ and $g^{-1}(F)$ are closed.
Since $f$ is continuous the set $f^{-1}(F)$ is a closed set in space $A$ which means that it can be written as $Pcap A$ where $P$ is a closed set in $Y$.
But that means that $f^{-1}(F)=Pcap A$ is an intersection of two closed sets: $A$ and $P$.
We conclude that $f^{-1}(F)$ is closed.
Similarly we find that $g^{-1}(F)$ is closed and we are ready.
You can apply that here on $A=[0,0.5]$ and $B=[0.5,1]$ wich are closed subsets of $Y:=[0,1]$.
answered Jan 5 at 14:13
drhabdrhab
103k545136
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add a comment |
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$begingroup$
It is also true more generally if $X$ is a topological space not necessarily metrizable.
In general if $A,B$ are closed subsets of topological space $Y$ and we have function $f:Ato X$ continuous on $A$, function $g:Bto X$ continuous on $B$ such that $f$ and $g$ coincide on $Acap B$ then the function $h=fcup g:Acup Bto X$ is continuous as well.
To prove that $h$ is continuous it is enough to show that the preimage of a closed set $Fsubseteq X$ is a closed subset of $Y$.
This preimage is the set $f^{-1}(F)cup g^{-1}(F)$, so we are ready if we can prove that the sets $f^{-1}(F)$ and $g^{-1}(F)$ are closed.
Since $f$ is continuous the set $f^{-1}(F)$ is a closed set in space $A$ which means that it can be written as $Pcap A$ where $P$ is a closed set in $Y$.
But that means that $f^{-1}(F)=Pcap A$ is an intersection of two closed sets: $A$ and $P$.
We conclude that $f^{-1}(F)$ is closed.
Similarly we find that $g^{-1}(F)$ is closed and we are ready.
You can apply that here on $A=[0,0.5]$ and $B=[0.5,1]$ wich are closed subsets of $Y:=[0,1]$.
answered Jan 5 at 14:13
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
It is also true more generally if $X$ is a topological space not necessarily metrizable.
In general if $A,B$ are closed subsets of topological space $Y$ and we have function $f:Ato X$ continuous on $A$, function $g:Bto X$ continuous on $B$ such that $f$ and $g$ coincide on $Acap B$ then the function $h=fcup g:Acup Bto X$ is continuous as well.
To prove that $h$ is continuous it is enough to show that the preimage of a closed set $Fsubseteq X$ is a closed subset of $Y$.
This preimage is the set $f^{-1}(F)cup g^{-1}(F)$, so we are ready if we can prove that the sets $f^{-1}(F)$ and $g^{-1}(F)$ are closed.
Since $f$ is continuous the set $f^{-1}(F)$ is a closed set in space $A$ which means that it can be written as $Pcap A$ where $P$ is a closed set in $Y$.
But that means that $f^{-1}(F)=Pcap A$ is an intersection of two closed sets: $A$ and $P$.
We conclude that $f^{-1}(F)$ is closed.
Similarly we find that $g^{-1}(F)$ is closed and we are ready.
You can apply that here on $A=[0,0.5]$ and $B=[0.5,1]$ wich are closed subsets of $Y:=[0,1]$.
answered Jan 5 at 14:13
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
It is also true more generally if $X$ is a topological space not necessarily metrizable.
In general if $A,B$ are closed subsets of topological space $Y$ and we have function $f:Ato X$ continuous on $A$, function $g:Bto X$ continuous on $B$ such that $f$ and $g$ coincide on $Acap B$ then the function $h=fcup g:Acup Bto X$ is continuous as well.
To prove that $h$ is continuous it is enough to show that the preimage of a closed set $Fsubseteq X$ is a closed subset of $Y$.
This preimage is the set $f^{-1}(F)cup g^{-1}(F)$, so we are ready if we can prove that the sets $f^{-1}(F)$ and $g^{-1}(F)$ are closed.
Since $f$ is continuous the set $f^{-1}(F)$ is a closed set in space $A$ which means that it can be written as $Pcap A$ where $P$ is a closed set in $Y$.
But that means that $f^{-1}(F)=Pcap A$ is an intersection of two closed sets: $A$ and $P$.
We conclude that $f^{-1}(F)$ is closed.
Similarly we find that $g^{-1}(F)$ is closed and we are ready.
You can apply that here on $A=[0,0.5]$ and $B=[0.5,1]$ wich are closed subsets of $Y:=[0,1]$.
answered Jan 5 at 14:13
drhabdrhab
103k545136
$endgroup$
It is also true more generally if $X$ is a topological space not necessarily metrizable.
In general if $A,B$ are closed subsets of topological space $Y$ and we have function $f:Ato X$ continuous on $A$, function $g:Bto X$ continuous on $B$ such that $f$ and $g$ coincide on $Acap B$ then the function $h=fcup g:Acup Bto X$ is continuous as well.
To prove that $h$ is continuous it is enough to show that the preimage of a closed set $Fsubseteq X$ is a closed subset of $Y$.
This preimage is the set $f^{-1}(F)cup g^{-1}(F)$, so we are ready if we can prove that the sets $f^{-1}(F)$ and $g^{-1}(F)$ are closed.
Since $f$ is continuous the set $f^{-1}(F)$ is a closed set in space $A$ which means that it can be written as $Pcap A$ where $P$ is a closed set in $Y$.
But that means that $f^{-1}(F)=Pcap A$ is an intersection of two closed sets: $A$ and $P$.
We conclude that $f^{-1}(F)$ is closed.
Similarly we find that $g^{-1}(F)$ is closed and we are ready.
You can apply that here on $A=[0,0.5]$ and $B=[0.5,1]$ wich are closed subsets of $Y:=[0,1]$.
answered Jan 5 at 14:13
drhabdrhab
103k545136
answered Jan 5 at 14:13
drhabdrhab
103k545136
answered Jan 5 at 14:13
drhabdrhab
103k545136
answered Jan 5 at 14:13
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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$begingroup$
Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
$endgroup$
– Yadati Kiran
Jan 5 at 14:07