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Applying topological definition of continuity
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I am a physics student and have just recently started studying about topology. I am trying to apply the topological definition of continuity in this function.
$$f:(0,2)rightarrow(0,1]cup(2,3)$$
$$f(x) = begincases
x, & text if x in (0,1]\
x+1, &text if x in (1,2)
endcases$$
I think that the co-domain of this function is not a topological space because the co-domain itself is not an open set and so it doesn't fit in the definition of a topology. So, is it so that we can't apply topological definition in this case or am I wrong somewhere?
general-topology
edited Jul 15 at 19:37
mechanodroid
22.3k52041
asked Jul 15 at 19:28
Yaman Sanghavi
1915
add a comment |Â
up vote
0
down vote
favorite
I am a physics student and have just recently started studying about topology. I am trying to apply the topological definition of continuity in this function.
$$f:(0,2)rightarrow(0,1]cup(2,3)$$
$$f(x) = begincases
x, & text if x in (0,1]\
x+1, &text if x in (1,2)
endcases$$
I think that the co-domain of this function is not a topological space because the co-domain itself is not an open set and so it doesn't fit in the definition of a topology. So, is it so that we can't apply topological definition in this case or am I wrong somewhere?
general-topology
edited Jul 15 at 19:37
mechanodroid
22.3k52041
asked Jul 15 at 19:28
Yaman Sanghavi
1915
Is $(0,1]$ an open set in the codomain? If so, then its inverse under $f$ must be an open set in $(0,2)$ in order for $f$ to be continuous.
– John Wayland Bales
Jul 15 at 19:36
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am a physics student and have just recently started studying about topology. I am trying to apply the topological definition of continuity in this function.
$$f:(0,2)rightarrow(0,1]cup(2,3)$$
$$f(x) = begincases
x, & text if x in (0,1]\
x+1, &text if x in (1,2)
endcases$$
I think that the co-domain of this function is not a topological space because the co-domain itself is not an open set and so it doesn't fit in the definition of a topology. So, is it so that we can't apply topological definition in this case or am I wrong somewhere?
general-topology
edited Jul 15 at 19:37
mechanodroid
22.3k52041
asked Jul 15 at 19:28
Yaman Sanghavi
1915
I am a physics student and have just recently started studying about topology. I am trying to apply the topological definition of continuity in this function.
$$f:(0,2)rightarrow(0,1]cup(2,3)$$
$$f(x) = begincases
x, & text if x in (0,1]\
x+1, &text if x in (1,2)
endcases$$
I think that the co-domain of this function is not a topological space because the co-domain itself is not an open set and so it doesn't fit in the definition of a topology. So, is it so that we can't apply topological definition in this case or am I wrong somewhere?
general-topology
edited Jul 15 at 19:37
mechanodroid
22.3k52041
asked Jul 15 at 19:28
Yaman Sanghavi
1915
edited Jul 15 at 19:37
mechanodroid
22.3k52041
edited Jul 15 at 19:37
mechanodroid
22.3k52041
edited Jul 15 at 19:37
mechanodroid
22.3k52041
22.3k52041
asked Jul 15 at 19:28
Yaman Sanghavi
1915
asked Jul 15 at 19:28
Yaman Sanghavi
1915
asked Jul 15 at 19:28
Yaman Sanghavi
1915
1915
Is $(0,1]$ an open set in the codomain? If so, then its inverse under $f$ must be an open set in $(0,2)$ in order for $f$ to be continuous.
– John Wayland Bales
Jul 15 at 19:36
add a comment |Â
Is $(0,1]$ an open set in the codomain? If so, then its inverse under $f$ must be an open set in $(0,2)$ in order for $f$ to be continuous.
– John Wayland Bales
Jul 15 at 19:36
Is $(0,1]$ an open set in the codomain? If so, then its inverse under $f$ must be an open set in $(0,2)$ in order for $f$ to be continuous.
– John Wayland Bales
Jul 15 at 19:36
Is $(0,1]$ an open set in the codomain? If so, then its inverse under $f$ must be an open set in $(0,2)$ in order for $f$ to be continuous.
– John Wayland Bales
Jul 15 at 19:36
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
The set $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$ is an open set in the codomain $(0,1] cup (2,3)$. We have
$$f^-1Big(left(0, 1right]Big) = left(0, 1right]$$
which is not an open set in $(0,2)$. Therefore $f$ is not continuous.
answered Jul 15 at 19:35
mechanodroid
22.3k52041
Why the downvote?
– copper.hat
Jul 15 at 19:50
1
I'm not the downvoter, but it might be because even though this answer is a valid proof of the discontinuity of the function is doesn't answer the questions asked by the OP.
– JonathanZ
Jul 15 at 19:52
1
I didn't downvote. I am still trying to understand the answers.
– Yaman Sanghavi
Jul 15 at 19:54
Why is this open $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$? I recall reading somewhere that we say that a set is open to the set it's contained in, not the larger metric space. If that's true then why is this set open in the codomain
– john fowles
Jul 15 at 20:41
1
@johnfowles Let $X$ be a topological space and $Y subseteq X$. A set $A subseteq Y$ is said to be open in $Y$ if there exists $U subseteq X$ open in $X$ such that $A = U cap Y$. In this case I took $X = mathbbR$, $Y = (0,1] cup (2,3)$, $A = (0,1]$ and $U = left(0, frac32right)$.
– mechanodroid
Jul 15 at 21:28
add a comment |Â
up vote
0
down vote
Every subset of a topological space inherits a default topology from its containing space, called, unsurprisingly, the "subspace topology". (You can look up this term on, say, Wikipedia, or in any topology text to read more details about it.) This is certainly the topology that is intended for $(0.1] cup (2,3)$ (considered as a subset of $mathbbR$ with its usual topology), and it does make it into a valid topological space.
But you'll have to be careful, as the definition of subspace topology can lead to some surprises for a first-time student. For instance, $(0,1]$ is an open subset of $(0.1] cup (2,3)$, even though it is not an open subset of $mathbbR$. (@mechanodroid uses this fact in their proof that the function is not continuous.)
answered Jul 15 at 19:48
JonathanZ
1,702412
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The set $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$ is an open set in the codomain $(0,1] cup (2,3)$. We have
$$f^-1Big(left(0, 1right]Big) = left(0, 1right]$$
which is not an open set in $(0,2)$. Therefore $f$ is not continuous.
answered Jul 15 at 19:35
mechanodroid
22.3k52041
Why the downvote?
– copper.hat
Jul 15 at 19:50
1
I'm not the downvoter, but it might be because even though this answer is a valid proof of the discontinuity of the function is doesn't answer the questions asked by the OP.
– JonathanZ
Jul 15 at 19:52
1
I didn't downvote. I am still trying to understand the answers.
– Yaman Sanghavi
Jul 15 at 19:54
Why is this open $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$? I recall reading somewhere that we say that a set is open to the set it's contained in, not the larger metric space. If that's true then why is this set open in the codomain
– john fowles
Jul 15 at 20:41
1
@johnfowles Let $X$ be a topological space and $Y subseteq X$. A set $A subseteq Y$ is said to be open in $Y$ if there exists $U subseteq X$ open in $X$ such that $A = U cap Y$. In this case I took $X = mathbbR$, $Y = (0,1] cup (2,3)$, $A = (0,1]$ and $U = left(0, frac32right)$.
– mechanodroid
Jul 15 at 21:28
add a comment |Â
up vote
3
down vote
The set $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$ is an open set in the codomain $(0,1] cup (2,3)$. We have
$$f^-1Big(left(0, 1right]Big) = left(0, 1right]$$
which is not an open set in $(0,2)$. Therefore $f$ is not continuous.
answered Jul 15 at 19:35
mechanodroid
22.3k52041
Why the downvote?
– copper.hat
Jul 15 at 19:50
1
I'm not the downvoter, but it might be because even though this answer is a valid proof of the discontinuity of the function is doesn't answer the questions asked by the OP.
– JonathanZ
Jul 15 at 19:52
1
I didn't downvote. I am still trying to understand the answers.
– Yaman Sanghavi
Jul 15 at 19:54
Why is this open $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$? I recall reading somewhere that we say that a set is open to the set it's contained in, not the larger metric space. If that's true then why is this set open in the codomain
– john fowles
Jul 15 at 20:41
1
@johnfowles Let $X$ be a topological space and $Y subseteq X$. A set $A subseteq Y$ is said to be open in $Y$ if there exists $U subseteq X$ open in $X$ such that $A = U cap Y$. In this case I took $X = mathbbR$, $Y = (0,1] cup (2,3)$, $A = (0,1]$ and $U = left(0, frac32right)$.
– mechanodroid
Jul 15 at 21:28
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The set $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$ is an open set in the codomain $(0,1] cup (2,3)$. We have
$$f^-1Big(left(0, 1right]Big) = left(0, 1right]$$
which is not an open set in $(0,2)$. Therefore $f$ is not continuous.
answered Jul 15 at 19:35
mechanodroid
22.3k52041
The set $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$ is an open set in the codomain $(0,1] cup (2,3)$. We have
$$f^-1Big(left(0, 1right]Big) = left(0, 1right]$$
which is not an open set in $(0,2)$. Therefore $f$ is not continuous.
answered Jul 15 at 19:35
mechanodroid
22.3k52041
answered Jul 15 at 19:35
mechanodroid
22.3k52041
answered Jul 15 at 19:35
mechanodroid
22.3k52041
answered Jul 15 at 19:35
mechanodroid
22.3k52041
22.3k52041
Why the downvote?
– copper.hat
Jul 15 at 19:50
1
I'm not the downvoter, but it might be because even though this answer is a valid proof of the discontinuity of the function is doesn't answer the questions asked by the OP.
– JonathanZ
Jul 15 at 19:52
1
I didn't downvote. I am still trying to understand the answers.
– Yaman Sanghavi
Jul 15 at 19:54
Why is this open $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$? I recall reading somewhere that we say that a set is open to the set it's contained in, not the larger metric space. If that's true then why is this set open in the codomain
– john fowles
Jul 15 at 20:41
1
@johnfowles Let $X$ be a topological space and $Y subseteq X$. A set $A subseteq Y$ is said to be open in $Y$ if there exists $U subseteq X$ open in $X$ such that $A = U cap Y$. In this case I took $X = mathbbR$, $Y = (0,1] cup (2,3)$, $A = (0,1]$ and $U = left(0, frac32right)$.
– mechanodroid
Jul 15 at 21:28
add a comment |Â
Why the downvote?
– copper.hat
Jul 15 at 19:50
1
I'm not the downvoter, but it might be because even though this answer is a valid proof of the discontinuity of the function is doesn't answer the questions asked by the OP.
– JonathanZ
Jul 15 at 19:52
1
I didn't downvote. I am still trying to understand the answers.
– Yaman Sanghavi
Jul 15 at 19:54
Why is this open $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$? I recall reading somewhere that we say that a set is open to the set it's contained in, not the larger metric space. If that's true then why is this set open in the codomain
– john fowles
Jul 15 at 20:41
1
@johnfowles Let $X$ be a topological space and $Y subseteq X$. A set $A subseteq Y$ is said to be open in $Y$ if there exists $U subseteq X$ open in $X$ such that $A = U cap Y$. In this case I took $X = mathbbR$, $Y = (0,1] cup (2,3)$, $A = (0,1]$ and $U = left(0, frac32right)$.
– mechanodroid
Jul 15 at 21:28
Why the downvote?
– copper.hat
Jul 15 at 19:50
Why the downvote?
– copper.hat
Jul 15 at 19:50
1
1
I'm not the downvoter, but it might be because even though this answer is a valid proof of the discontinuity of the function is doesn't answer the questions asked by the OP.
– JonathanZ
Jul 15 at 19:52
I'm not the downvoter, but it might be because even though this answer is a valid proof of the discontinuity of the function is doesn't answer the questions asked by the OP.
– JonathanZ
Jul 15 at 19:52
1
1
I didn't downvote. I am still trying to understand the answers.
– Yaman Sanghavi
Jul 15 at 19:54
I didn't downvote. I am still trying to understand the answers.
– Yaman Sanghavi
Jul 15 at 19:54
Why is this open $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$? I recall reading somewhere that we say that a set is open to the set it's contained in, not the larger metric space. If that's true then why is this set open in the codomain
– john fowles
Jul 15 at 20:41
Why is this open $left(0, 1right] = left(0, frac32right) cap Big((0,1] cup (2,3)Big)$? I recall reading somewhere that we say that a set is open to the set it's contained in, not the larger metric space. If that's true then why is this set open in the codomain
– john fowles
Jul 15 at 20:41
1
1
@johnfowles Let $X$ be a topological space and $Y subseteq X$. A set $A subseteq Y$ is said to be open in $Y$ if there exists $U subseteq X$ open in $X$ such that $A = U cap Y$. In this case I took $X = mathbbR$, $Y = (0,1] cup (2,3)$, $A = (0,1]$ and $U = left(0, frac32right)$.
– mechanodroid
Jul 15 at 21:28
@johnfowles Let $X$ be a topological space and $Y subseteq X$. A set $A subseteq Y$ is said to be open in $Y$ if there exists $U subseteq X$ open in $X$ such that $A = U cap Y$. In this case I took $X = mathbbR$, $Y = (0,1] cup (2,3)$, $A = (0,1]$ and $U = left(0, frac32right)$.
– mechanodroid
Jul 15 at 21:28
add a comment |Â
up vote
0
down vote
Every subset of a topological space inherits a default topology from its containing space, called, unsurprisingly, the "subspace topology". (You can look up this term on, say, Wikipedia, or in any topology text to read more details about it.) This is certainly the topology that is intended for $(0.1] cup (2,3)$ (considered as a subset of $mathbbR$ with its usual topology), and it does make it into a valid topological space.
But you'll have to be careful, as the definition of subspace topology can lead to some surprises for a first-time student. For instance, $(0,1]$ is an open subset of $(0.1] cup (2,3)$, even though it is not an open subset of $mathbbR$. (@mechanodroid uses this fact in their proof that the function is not continuous.)
answered Jul 15 at 19:48
JonathanZ
1,702412
add a comment |Â
up vote
0
down vote
Every subset of a topological space inherits a default topology from its containing space, called, unsurprisingly, the "subspace topology". (You can look up this term on, say, Wikipedia, or in any topology text to read more details about it.) This is certainly the topology that is intended for $(0.1] cup (2,3)$ (considered as a subset of $mathbbR$ with its usual topology), and it does make it into a valid topological space.
But you'll have to be careful, as the definition of subspace topology can lead to some surprises for a first-time student. For instance, $(0,1]$ is an open subset of $(0.1] cup (2,3)$, even though it is not an open subset of $mathbbR$. (@mechanodroid uses this fact in their proof that the function is not continuous.)
answered Jul 15 at 19:48
JonathanZ
1,702412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Every subset of a topological space inherits a default topology from its containing space, called, unsurprisingly, the "subspace topology". (You can look up this term on, say, Wikipedia, or in any topology text to read more details about it.) This is certainly the topology that is intended for $(0.1] cup (2,3)$ (considered as a subset of $mathbbR$ with its usual topology), and it does make it into a valid topological space.
But you'll have to be careful, as the definition of subspace topology can lead to some surprises for a first-time student. For instance, $(0,1]$ is an open subset of $(0.1] cup (2,3)$, even though it is not an open subset of $mathbbR$. (@mechanodroid uses this fact in their proof that the function is not continuous.)
answered Jul 15 at 19:48
JonathanZ
1,702412
Every subset of a topological space inherits a default topology from its containing space, called, unsurprisingly, the "subspace topology". (You can look up this term on, say, Wikipedia, or in any topology text to read more details about it.) This is certainly the topology that is intended for $(0.1] cup (2,3)$ (considered as a subset of $mathbbR$ with its usual topology), and it does make it into a valid topological space.
But you'll have to be careful, as the definition of subspace topology can lead to some surprises for a first-time student. For instance, $(0,1]$ is an open subset of $(0.1] cup (2,3)$, even though it is not an open subset of $mathbbR$. (@mechanodroid uses this fact in their proof that the function is not continuous.)
answered Jul 15 at 19:48
JonathanZ
1,702412
edited Jul 15 at 19:53
answered Jul 15 at 19:48
JonathanZ
1,702412
answered Jul 15 at 19:48
JonathanZ
1,702412
answered Jul 15 at 19:48
JonathanZ
1,702412
1,702412
add a comment |Â
add a comment |Â
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Is $(0,1]$ an open set in the codomain? If so, then its inverse under $f$ must be an open set in $(0,2)$ in order for $f$ to be continuous.
– John Wayland Bales
Jul 15 at 19:36