
What is the boundedness theorem?
Answer
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Hint: We are asked to explain the boundedness theorem. Boundedness theorem in simple words can be stated as a continuous function which is defined on a closed interval and it has an upper bound to it. We can carry out the proof for validity of this theorem by using contradiction. Hence, we will have the boundedness theorem.
Complete step-by-step solution:
According to the given question, we are asked to write about and explain about the boundedness theorem.
Boundedness theorem states that if there is a function ‘f’ and it is continuous and is defined on a closed interval \[[a,b]\], then the given function ‘f’ is bounded in that interval.
A continuous function refers to a function with no discontinuities or in other words no abrupt changes in the values.
And a closed interval refers to the interval whose endpoints are also included and it is denoted by square brackets.
We can prove this theorem by contradiction,
Let there be a function \[f(x)\] which is continuous in nature and is defined on the closed interval \[[a,b]\] but without an upper bound.
Then, for all \[n\in N\], we have \[{{x}_{n}}\in [a,b]\] such that the \[f({{x}_{n}})>n\].
Now, let there be a point \[c\] in the interval \[[a,b]\] where \[{{x}_{n}}\] is dense.
Since we have the sequence of \[{{x}_{n}}\] dense at \[c\], we have an increasing sequence \[{{n}_{k}}\in N\] such that, \[{{x}_{{{n}_{k}}}}\to c\] as \[k \to \infty \].
We know that, since \[f(x)\] is continuous at \[c\], so we have,
\[\displaystyle \lim_{x \to c}f(x)=f(c)\] and it is bounded.
But,
\[\displaystyle \lim_{k \to \infty }{{x}_{{{n}_{k}}}}=c\] and \[\displaystyle \lim_{k \to \infty }f({{x}_{{{n}_{k}}}})=\infty \]
Which is not bounded and we have a contradiction here.
So, there is no such \[f(x)\] without the upper bound.
Note: The boundedness theorem is only valid for continuous function because for a function with discontinuity, we will have no value of \[f(x)\] which is not in accordance with the boundedness theorem. Also, in the given closed interval for every value of \[{{x}_{n}}\], we have the value of \[f({{x}_{n}})>n\].
Complete step-by-step solution:
According to the given question, we are asked to write about and explain about the boundedness theorem.
Boundedness theorem states that if there is a function ‘f’ and it is continuous and is defined on a closed interval \[[a,b]\], then the given function ‘f’ is bounded in that interval.
A continuous function refers to a function with no discontinuities or in other words no abrupt changes in the values.
And a closed interval refers to the interval whose endpoints are also included and it is denoted by square brackets.
We can prove this theorem by contradiction,
Let there be a function \[f(x)\] which is continuous in nature and is defined on the closed interval \[[a,b]\] but without an upper bound.
Then, for all \[n\in N\], we have \[{{x}_{n}}\in [a,b]\] such that the \[f({{x}_{n}})>n\].
Now, let there be a point \[c\] in the interval \[[a,b]\] where \[{{x}_{n}}\] is dense.
Since we have the sequence of \[{{x}_{n}}\] dense at \[c\], we have an increasing sequence \[{{n}_{k}}\in N\] such that, \[{{x}_{{{n}_{k}}}}\to c\] as \[k \to \infty \].
We know that, since \[f(x)\] is continuous at \[c\], so we have,
\[\displaystyle \lim_{x \to c}f(x)=f(c)\] and it is bounded.
But,
\[\displaystyle \lim_{k \to \infty }{{x}_{{{n}_{k}}}}=c\] and \[\displaystyle \lim_{k \to \infty }f({{x}_{{{n}_{k}}}})=\infty \]
Which is not bounded and we have a contradiction here.
So, there is no such \[f(x)\] without the upper bound.
Note: The boundedness theorem is only valid for continuous function because for a function with discontinuity, we will have no value of \[f(x)\] which is not in accordance with the boundedness theorem. Also, in the given closed interval for every value of \[{{x}_{n}}\], we have the value of \[f({{x}_{n}})>n\].
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