Every Interval $I subset mathbb{R}$ is connected
The proof I've been given is
If not, there is a non-constant continuous $f$ from $I$ to
discrete ${0, 1}$. Then $f: I rightarrow mathbb{R}$ is also continuous which contradicts the Intermediate Value Theorem.
Why must $f: I rightarrow mathbb{R}$ also be continuous?
general-topology
edited Dec 9 at 0:19
amWhy
191k28224439
asked May 20 '17 at 15:44
user440780
143
add a comment |
The proof I've been given is
If not, there is a non-constant continuous $f$ from $I$ to
discrete ${0, 1}$. Then $f: I rightarrow mathbb{R}$ is also continuous which contradicts the Intermediate Value Theorem.
Why must $f: I rightarrow mathbb{R}$ also be continuous?
general-topology
edited Dec 9 at 0:19
amWhy
191k28224439
asked May 20 '17 at 15:44
user440780
143
$f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
– fleablood
May 20 '17 at 15:58
1
I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
– fleablood
May 20 '17 at 15:59
1
It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
– Matematleta
May 20 '17 at 16:02
add a comment |
The proof I've been given is
If not, there is a non-constant continuous $f$ from $I$ to
discrete ${0, 1}$. Then $f: I rightarrow mathbb{R}$ is also continuous which contradicts the Intermediate Value Theorem.
Why must $f: I rightarrow mathbb{R}$ also be continuous?
general-topology
edited Dec 9 at 0:19
amWhy
191k28224439
asked May 20 '17 at 15:44
user440780
143
The proof I've been given is
If not, there is a non-constant continuous $f$ from $I$ to
discrete ${0, 1}$. Then $f: I rightarrow mathbb{R}$ is also continuous which contradicts the Intermediate Value Theorem.
Why must $f: I rightarrow mathbb{R}$ also be continuous?
general-topology
general-topology
edited Dec 9 at 0:19
amWhy
191k28224439
asked May 20 '17 at 15:44
user440780
143
edited Dec 9 at 0:19
amWhy
191k28224439
asked May 20 '17 at 15:44
user440780
143
edited Dec 9 at 0:19
amWhy
191k28224439
edited Dec 9 at 0:19
amWhy
191k28224439
edited Dec 9 at 0:19
amWhy
191k28224439
191k28224439
asked May 20 '17 at 15:44
user440780
143
asked May 20 '17 at 15:44
user440780
143
asked May 20 '17 at 15:44
user440780
143
143
$f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
– fleablood
May 20 '17 at 15:58
1
I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
– fleablood
May 20 '17 at 15:59
1
It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
– Matematleta
May 20 '17 at 16:02
add a comment |
$f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
– fleablood
May 20 '17 at 15:58
1
I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
– fleablood
May 20 '17 at 15:59
1
It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
– Matematleta
May 20 '17 at 16:02
$f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
– fleablood
May 20 '17 at 15:58
$f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
– fleablood
May 20 '17 at 15:58
1
1
I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
– fleablood
May 20 '17 at 15:59
I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
– fleablood
May 20 '17 at 15:59
1
1
It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
– Matematleta
May 20 '17 at 16:02
It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
– Matematleta
May 20 '17 at 16:02
add a comment |
2 Answers
2
active
oldest
votes
The inclusion ${0,1} to mathbb{R}$ is continuous because ${0,1}$ has the discrete topology.
Compositions of continuous functions are continuous.
answered May 20 '17 at 15:47
user133281
13.6k22551
add a comment |
If $I$ is not connected, then by definition there are $A$ and $B$ both open such that $A cup B=I$, if you consider $f$ such that $f(A)=0$ and $f(B)=1$ than this is a continuos function by definition, infact in $ {0,1} $ as a subset of $mathbb R$ there is the discrete topology and $ f^{-1}( {0,1})=I $ that is open, $ f^{-1}( {0})=A $ that is open, $ f^{-1}( {1})=B $ that is open and
$ f^{-1}(emptyset)=emptyset $ that is open so you have a contraddiction with the IVT. However this demonstration is circular because you have to use the connectedness of $I$ for proving the IVT.
answered May 20 '17 at 15:56
StabiloBoss
865
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
The inclusion ${0,1} to mathbb{R}$ is continuous because ${0,1}$ has the discrete topology.
Compositions of continuous functions are continuous.
answered May 20 '17 at 15:47
user133281
13.6k22551
add a comment |
The inclusion ${0,1} to mathbb{R}$ is continuous because ${0,1}$ has the discrete topology.
Compositions of continuous functions are continuous.
answered May 20 '17 at 15:47
user133281
13.6k22551
add a comment |
The inclusion ${0,1} to mathbb{R}$ is continuous because ${0,1}$ has the discrete topology.
Compositions of continuous functions are continuous.
answered May 20 '17 at 15:47
user133281
13.6k22551
The inclusion ${0,1} to mathbb{R}$ is continuous because ${0,1}$ has the discrete topology.
Compositions of continuous functions are continuous.
answered May 20 '17 at 15:47
user133281
13.6k22551
answered May 20 '17 at 15:47
user133281
13.6k22551
answered May 20 '17 at 15:47
user133281
13.6k22551
answered May 20 '17 at 15:47
user133281
13.6k22551
13.6k22551
add a comment |
add a comment |
If $I$ is not connected, then by definition there are $A$ and $B$ both open such that $A cup B=I$, if you consider $f$ such that $f(A)=0$ and $f(B)=1$ than this is a continuos function by definition, infact in $ {0,1} $ as a subset of $mathbb R$ there is the discrete topology and $ f^{-1}( {0,1})=I $ that is open, $ f^{-1}( {0})=A $ that is open, $ f^{-1}( {1})=B $ that is open and
$ f^{-1}(emptyset)=emptyset $ that is open so you have a contraddiction with the IVT. However this demonstration is circular because you have to use the connectedness of $I$ for proving the IVT.
answered May 20 '17 at 15:56
StabiloBoss
865
add a comment |
If $I$ is not connected, then by definition there are $A$ and $B$ both open such that $A cup B=I$, if you consider $f$ such that $f(A)=0$ and $f(B)=1$ than this is a continuos function by definition, infact in $ {0,1} $ as a subset of $mathbb R$ there is the discrete topology and $ f^{-1}( {0,1})=I $ that is open, $ f^{-1}( {0})=A $ that is open, $ f^{-1}( {1})=B $ that is open and
$ f^{-1}(emptyset)=emptyset $ that is open so you have a contraddiction with the IVT. However this demonstration is circular because you have to use the connectedness of $I$ for proving the IVT.
answered May 20 '17 at 15:56
StabiloBoss
865
add a comment |
If $I$ is not connected, then by definition there are $A$ and $B$ both open such that $A cup B=I$, if you consider $f$ such that $f(A)=0$ and $f(B)=1$ than this is a continuos function by definition, infact in $ {0,1} $ as a subset of $mathbb R$ there is the discrete topology and $ f^{-1}( {0,1})=I $ that is open, $ f^{-1}( {0})=A $ that is open, $ f^{-1}( {1})=B $ that is open and
$ f^{-1}(emptyset)=emptyset $ that is open so you have a contraddiction with the IVT. However this demonstration is circular because you have to use the connectedness of $I$ for proving the IVT.
answered May 20 '17 at 15:56
StabiloBoss
865
If $I$ is not connected, then by definition there are $A$ and $B$ both open such that $A cup B=I$, if you consider $f$ such that $f(A)=0$ and $f(B)=1$ than this is a continuos function by definition, infact in $ {0,1} $ as a subset of $mathbb R$ there is the discrete topology and $ f^{-1}( {0,1})=I $ that is open, $ f^{-1}( {0})=A $ that is open, $ f^{-1}( {1})=B $ that is open and
$ f^{-1}(emptyset)=emptyset $ that is open so you have a contraddiction with the IVT. However this demonstration is circular because you have to use the connectedness of $I$ for proving the IVT.
answered May 20 '17 at 15:56
StabiloBoss
865
edited May 20 '17 at 16:36
answered May 20 '17 at 15:56
StabiloBoss
865
answered May 20 '17 at 15:56
StabiloBoss
865
answered May 20 '17 at 15:56
StabiloBoss
865
865
add a comment |
add a comment |
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$f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
– fleablood
May 20 '17 at 15:58
1
I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
– fleablood
May 20 '17 at 15:59
1
It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
– Matematleta
May 20 '17 at 16:02