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September 20th, 2018, 02:35 PM   #1
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f bounded on A and B, bounded on A U B?

Prove that if f is bounded on A and f is also bounded on B, then f is bounded on A U B

Proof

Since $\displaystyle f$ is bounded on A, there is a real number $\displaystyle m$ such that

$\displaystyle \left|f(x) \right| \leq m$, for all x in A.

Similarly, since $\displaystyle f$ is bounded on B, there is an $\displaystyle n$ such that

$\displaystyle \left|f(x) \right| \leq n$, for all x in B.

Let $\displaystyle M = \max(m,n)$ then

$\displaystyle \left|f(x) \right| \leq M$, for all x in $\displaystyle A \cup B$.

Thus showing that $\displaystyle f$ is bounded on $\displaystyle A \cup B$

Is this proof right?

Last edited by skipjack; November 7th, 2018 at 09:47 PM.
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September 20th, 2018, 03:46 PM   #2
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I would like to see a little more detail. In particular, could you explain why it follows that $\displaystyle \left|f(x) \right| \leq M$, for all x in $\displaystyle A \cup B$?
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September 21st, 2018, 11:26 AM   #3
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Does this work?

n,m are positive real numbers. So either $\displaystyle n=m,\ n<m, $ or $\displaystyle m<n$.
If $\displaystyle n=m$ then

$\displaystyle |f(x)| \leq n=m $ For all x in A
$\displaystyle |f(x)| \leq n=m $ For all x in b

Thus for either set $\displaystyle n=m$ is a bound, and so, $\displaystyle f$ is bounded on $\displaystyle A \cup B$.

If $\displaystyle n < m$ then

$\displaystyle |f(x)| \leq m $ For all x in A
$\displaystyle |f(x)| \leq n<m $ For all x in B

Thus for either set $\displaystyle m$ is a bound, and so, $\displaystyle f$ is bounded on $\displaystyle A \cup B$

If $\displaystyle m < n$ then

$\displaystyle |f(x)| \leq m<n $ For all x in A
$\displaystyle |f(x)| \leq n $ For all x in B

Thus for either set $\displaystyle n$ is a bound, and so, $\displaystyle f$ is bounded on $\displaystyle A \cup B$

So if $\displaystyle f$ is bounded on sets A and B, then $\displaystyle f$ is bounded on there union $\displaystyle A \cup B$, which is what I needed to show. I hope I got something right. Thanks for responding to post. If I missed something or everything, let me know.

Last edited by skipjack; November 7th, 2018 at 09:49 PM.
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November 7th, 2018, 09:52 PM   #4
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Both versions look okay to me.
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