Continuous function maps connected set onto connected set
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Let $X, Y$ be topological spaces and $f: Xto Y$ continuous. Show that $f$ maps connected sets onto connected sets.
I am not sure if I understand the definition of a connected set right.
Our definitions are as follows:
A topological space $X$ is "connected" if $emptyset$ and $X$ are the only sets, which are closed and open in $X$.
A subset of a topological space is "connected" if it is "connected" with regards to the subspace topology.
Therefor a set $Asubseteq X$ is connected, if $A$ and $emptyset$ are the only closed and open ("clopen") sets with regards to the subspace topology?
Now for the proof:
Let $emptysetneq Asubset X$ be connected. Therefor $A$ is not clopen with regards to the subspace topology.
I have to show, that $f(A)$ is not clopen with regards to the subspace topology.
Suppose $emptysetneq f(A)subset Y$ is clopen. Since $f$ is continuous the preimage of open and closed sets is open and closed. Hence $A$ is clopen. Which contradicts the assumption.
Is this correct?
Thanks in advance.
general-topology proof-verification
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Let $X, Y$ be topological spaces and $f: Xto Y$ continuous. Show that $f$ maps connected sets onto connected sets.
I am not sure if I understand the definition of a connected set right.
Our definitions are as follows:
A topological space $X$ is "connected" if $emptyset$ and $X$ are the only sets, which are closed and open in $X$.
A subset of a topological space is "connected" if it is "connected" with regards to the subspace topology.
Therefor a set $Asubseteq X$ is connected, if $A$ and $emptyset$ are the only closed and open ("clopen") sets with regards to the subspace topology?
Now for the proof:
Let $emptysetneq Asubset X$ be connected. Therefor $A$ is not clopen with regards to the subspace topology.
I have to show, that $f(A)$ is not clopen with regards to the subspace topology.
Suppose $emptysetneq f(A)subset Y$ is clopen. Since $f$ is continuous the preimage of open and closed sets is open and closed. Hence $A$ is clopen. Which contradicts the assumption.
Is this correct?
Thanks in advance.
general-topology proof-verification
In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
â user583185
Aug 13 at 10:57
So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
â Cornman
Aug 13 at 10:57
@rTSlines My comment above is refering to your first comment. Was it wrong?
â Cornman
Aug 13 at 10:58
Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
â user583185
Aug 13 at 10:58
So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
â user583185
Aug 13 at 11:00
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Let $X, Y$ be topological spaces and $f: Xto Y$ continuous. Show that $f$ maps connected sets onto connected sets.
I am not sure if I understand the definition of a connected set right.
Our definitions are as follows:
A topological space $X$ is "connected" if $emptyset$ and $X$ are the only sets, which are closed and open in $X$.
A subset of a topological space is "connected" if it is "connected" with regards to the subspace topology.
Therefor a set $Asubseteq X$ is connected, if $A$ and $emptyset$ are the only closed and open ("clopen") sets with regards to the subspace topology?
Now for the proof:
Let $emptysetneq Asubset X$ be connected. Therefor $A$ is not clopen with regards to the subspace topology.
I have to show, that $f(A)$ is not clopen with regards to the subspace topology.
Suppose $emptysetneq f(A)subset Y$ is clopen. Since $f$ is continuous the preimage of open and closed sets is open and closed. Hence $A$ is clopen. Which contradicts the assumption.
Is this correct?
Thanks in advance.
general-topology proof-verification
Let $X, Y$ be topological spaces and $f: Xto Y$ continuous. Show that $f$ maps connected sets onto connected sets.
I am not sure if I understand the definition of a connected set right.
Our definitions are as follows:
A topological space $X$ is "connected" if $emptyset$ and $X$ are the only sets, which are closed and open in $X$.
A subset of a topological space is "connected" if it is "connected" with regards to the subspace topology.
Therefor a set $Asubseteq X$ is connected, if $A$ and $emptyset$ are the only closed and open ("clopen") sets with regards to the subspace topology?
Now for the proof:
Let $emptysetneq Asubset X$ be connected. Therefor $A$ is not clopen with regards to the subspace topology.
I have to show, that $f(A)$ is not clopen with regards to the subspace topology.
Suppose $emptysetneq f(A)subset Y$ is clopen. Since $f$ is continuous the preimage of open and closed sets is open and closed. Hence $A$ is clopen. Which contradicts the assumption.
Is this correct?
Thanks in advance.
general-topology proof-verification
edited Aug 13 at 10:49
asked Aug 13 at 10:38
Cornman
2,61921128
2,61921128
In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
â user583185
Aug 13 at 10:57
So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
â Cornman
Aug 13 at 10:57
@rTSlines My comment above is refering to your first comment. Was it wrong?
â Cornman
Aug 13 at 10:58
Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
â user583185
Aug 13 at 10:58
So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
â user583185
Aug 13 at 11:00
add a comment |Â
In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
â user583185
Aug 13 at 10:57
So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
â Cornman
Aug 13 at 10:57
@rTSlines My comment above is refering to your first comment. Was it wrong?
â Cornman
Aug 13 at 10:58
Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
â user583185
Aug 13 at 10:58
So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
â user583185
Aug 13 at 11:00
In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
â user583185
Aug 13 at 10:57
In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
â user583185
Aug 13 at 10:57
So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
â Cornman
Aug 13 at 10:57
So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
â Cornman
Aug 13 at 10:57
@rTSlines My comment above is refering to your first comment. Was it wrong?
â Cornman
Aug 13 at 10:58
@rTSlines My comment above is refering to your first comment. Was it wrong?
â Cornman
Aug 13 at 10:58
Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
â user583185
Aug 13 at 10:58
Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
â user583185
Aug 13 at 10:58
So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
â user583185
Aug 13 at 11:00
So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
â user583185
Aug 13 at 11:00
add a comment |Â
1 Answer
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You are saying that $Asubseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.
Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).
This comes to the same as proving that no set $Usubset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $Unotinvarnothing,f(A)$.
Suppose that there is such a set.
Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $Vnotinvarnothing, A$.
Proved is then: $f(A)$ not connected $implies A$ not connected, or equivalently:
$$Atext connected implies f(A)text connected$$
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You are saying that $Asubseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.
Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).
This comes to the same as proving that no set $Usubset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $Unotinvarnothing,f(A)$.
Suppose that there is such a set.
Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $Vnotinvarnothing, A$.
Proved is then: $f(A)$ not connected $implies A$ not connected, or equivalently:
$$Atext connected implies f(A)text connected$$
add a comment |Â
up vote
1
down vote
accepted
You are saying that $Asubseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.
Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).
This comes to the same as proving that no set $Usubset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $Unotinvarnothing,f(A)$.
Suppose that there is such a set.
Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $Vnotinvarnothing, A$.
Proved is then: $f(A)$ not connected $implies A$ not connected, or equivalently:
$$Atext connected implies f(A)text connected$$
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are saying that $Asubseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.
Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).
This comes to the same as proving that no set $Usubset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $Unotinvarnothing,f(A)$.
Suppose that there is such a set.
Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $Vnotinvarnothing, A$.
Proved is then: $f(A)$ not connected $implies A$ not connected, or equivalently:
$$Atext connected implies f(A)text connected$$
You are saying that $Asubseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.
Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).
This comes to the same as proving that no set $Usubset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $Unotinvarnothing,f(A)$.
Suppose that there is such a set.
Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $Vnotinvarnothing, A$.
Proved is then: $f(A)$ not connected $implies A$ not connected, or equivalently:
$$Atext connected implies f(A)text connected$$
answered Aug 13 at 10:58
drhab
87.4k541118
87.4k541118
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In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
â user583185
Aug 13 at 10:57
So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
â Cornman
Aug 13 at 10:57
@rTSlines My comment above is refering to your first comment. Was it wrong?
â Cornman
Aug 13 at 10:58
Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
â user583185
Aug 13 at 10:58
So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
â user583185
Aug 13 at 11:00