Mixing
Direct Proof that Continuous Functions Satisfy Epsilon-Delta Condition
$begingroup$
If we define a continuous real function $f: M to R$ as one that preserves sequential convergence:
$$
(x_n)to p Rightarrow f(x_n) to f(p), tag{1}
$$
then to show that this definition is equivalent to the epsilon-delta condition,
$$
forall p forall epsilon exists delta: x in V_{delta}(p) Rightarrow f(x) in V_{epsilon}(f(p)), tag{2}
$$
my textbooks use use a proof by contradiction in which they derive a convergent sequence that does not converge under the mapping.
Question: Is it possible to have a direct proof of $(1) to (2)$?
Thought: I think to directly show that $(1) to (2)$, we will have to show that every element of M is an element of at least a sequence in $M$, i.e., the union of the elements of all the sequences in $M$ is $M$.
real-analysis analysis
edited Jan 12 at 8:18
md2perpe
7,96111028
asked Jan 12 at 2:14
A Slow LearnerA Slow Learner
441212
$endgroup$
add a comment |
$begingroup$
If we define a continuous real function $f: M to R$ as one that preserves sequential convergence:
$$
(x_n)to p Rightarrow f(x_n) to f(p), tag{1}
$$
then to show that this definition is equivalent to the epsilon-delta condition,
$$
forall p forall epsilon exists delta: x in V_{delta}(p) Rightarrow f(x) in V_{epsilon}(f(p)), tag{2}
$$
my textbooks use use a proof by contradiction in which they derive a convergent sequence that does not converge under the mapping.
Question: Is it possible to have a direct proof of $(1) to (2)$?
Thought: I think to directly show that $(1) to (2)$, we will have to show that every element of M is an element of at least a sequence in $M$, i.e., the union of the elements of all the sequences in $M$ is $M$.
real-analysis analysis
edited Jan 12 at 8:18
md2perpe
7,96111028
asked Jan 12 at 2:14
A Slow LearnerA Slow Learner
441212
$endgroup$
$begingroup$
It's worth noting that sequential continuity can be weaker than continuity in a general topological space. I'm guessing $M$ is a metric space?
$endgroup$
– Theo Bendit
Jan 12 at 2:45
add a comment |
$begingroup$
If we define a continuous real function $f: M to R$ as one that preserves sequential convergence:
$$
(x_n)to p Rightarrow f(x_n) to f(p), tag{1}
$$
then to show that this definition is equivalent to the epsilon-delta condition,
$$
forall p forall epsilon exists delta: x in V_{delta}(p) Rightarrow f(x) in V_{epsilon}(f(p)), tag{2}
$$
my textbooks use use a proof by contradiction in which they derive a convergent sequence that does not converge under the mapping.
Question: Is it possible to have a direct proof of $(1) to (2)$?
Thought: I think to directly show that $(1) to (2)$, we will have to show that every element of M is an element of at least a sequence in $M$, i.e., the union of the elements of all the sequences in $M$ is $M$.
real-analysis analysis
edited Jan 12 at 8:18
md2perpe
7,96111028
asked Jan 12 at 2:14
A Slow LearnerA Slow Learner
441212
$endgroup$
If we define a continuous real function $f: M to R$ as one that preserves sequential convergence:
$$
(x_n)to p Rightarrow f(x_n) to f(p), tag{1}
$$
then to show that this definition is equivalent to the epsilon-delta condition,
$$
forall p forall epsilon exists delta: x in V_{delta}(p) Rightarrow f(x) in V_{epsilon}(f(p)), tag{2}
$$
my textbooks use use a proof by contradiction in which they derive a convergent sequence that does not converge under the mapping.
Question: Is it possible to have a direct proof of $(1) to (2)$?
Thought: I think to directly show that $(1) to (2)$, we will have to show that every element of M is an element of at least a sequence in $M$, i.e., the union of the elements of all the sequences in $M$ is $M$.
real-analysis analysis
real-analysis analysis
edited Jan 12 at 8:18
md2perpe
7,96111028
asked Jan 12 at 2:14
A Slow LearnerA Slow Learner
441212
edited Jan 12 at 8:18
md2perpe
7,96111028
asked Jan 12 at 2:14
A Slow LearnerA Slow Learner
441212
edited Jan 12 at 8:18
md2perpe
7,96111028
edited Jan 12 at 8:18
md2perpe
7,96111028
edited Jan 12 at 8:18
md2perpe
7,96111028
7,96111028
asked Jan 12 at 2:14
A Slow LearnerA Slow Learner
441212
asked Jan 12 at 2:14
A Slow LearnerA Slow Learner
441212
asked Jan 12 at 2:14
A Slow LearnerA Slow Learner
441212
441212
$begingroup$
It's worth noting that sequential continuity can be weaker than continuity in a general topological space. I'm guessing $M$ is a metric space?
$endgroup$
– Theo Bendit
Jan 12 at 2:45
add a comment |
$begingroup$
It's worth noting that sequential continuity can be weaker than continuity in a general topological space. I'm guessing $M$ is a metric space?
$endgroup$
– Theo Bendit
Jan 12 at 2:45
$begingroup$
It's worth noting that sequential continuity can be weaker than continuity in a general topological space. I'm guessing $M$ is a metric space?
$endgroup$
– Theo Bendit
Jan 12 at 2:45
$begingroup$
It's worth noting that sequential continuity can be weaker than continuity in a general topological space. I'm guessing $M$ is a metric space?
$endgroup$
– Theo Bendit
Jan 12 at 2:45
add a comment |
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$begingroup$
It's worth noting that sequential continuity can be weaker than continuity in a general topological space. I'm guessing $M$ is a metric space?
$endgroup$
– Theo Bendit
Jan 12 at 2:45