Mixing
If $X$ is a real normed linear space, then $B(0,r)=r B(0,1)$
$begingroup$
If $X$ is a real normed linear space, I want to prove that $$B_{r}(0)=r B_{1}(0)$$
My proof
Let $zin rB(0,1)$, there exists $xin B(0,1)$ such that $z=rx$.
$$|z-0|=|rx-0|=r|x-0|<r$$
So, $zin B(0,r)$ and $$rB(0,1)subseteq B(0,r).$$
On the other hand, let $zin B(0,r)$, then $|z-0|<r$. This implies that
$$bigg|dfrac{z}{r}-0bigg|<1.$$
Thus, $dfrac{z}{r}in B(0,1)$ and there exists $xin B(0,1)$ such that
$$dfrac{z}{r}=xiff z=rx in rB(0,1)$$
Hence, $$ B(0,r) subseteq rB(0,1).$$
Therefore,
$$ B(0,r) = rB(0,1).$$
Question
Kindly help check if my proof is correct.
general-topology functional-analysis metric-spaces
asked Jan 18 at 11:30
Omojola MichealOmojola Micheal
1,918324
$endgroup$
add a comment |
$begingroup$
If $X$ is a real normed linear space, I want to prove that $$B_{r}(0)=r B_{1}(0)$$
My proof
Let $zin rB(0,1)$, there exists $xin B(0,1)$ such that $z=rx$.
$$|z-0|=|rx-0|=r|x-0|<r$$
So, $zin B(0,r)$ and $$rB(0,1)subseteq B(0,r).$$
On the other hand, let $zin B(0,r)$, then $|z-0|<r$. This implies that
$$bigg|dfrac{z}{r}-0bigg|<1.$$
Thus, $dfrac{z}{r}in B(0,1)$ and there exists $xin B(0,1)$ such that
$$dfrac{z}{r}=xiff z=rx in rB(0,1)$$
Hence, $$ B(0,r) subseteq rB(0,1).$$
Therefore,
$$ B(0,r) = rB(0,1).$$
Question
Kindly help check if my proof is correct.
general-topology functional-analysis metric-spaces
asked Jan 18 at 11:30
Omojola MichealOmojola Micheal
1,918324
$endgroup$
add a comment |
$begingroup$
If $X$ is a real normed linear space, I want to prove that $$B_{r}(0)=r B_{1}(0)$$
My proof
Let $zin rB(0,1)$, there exists $xin B(0,1)$ such that $z=rx$.
$$|z-0|=|rx-0|=r|x-0|<r$$
So, $zin B(0,r)$ and $$rB(0,1)subseteq B(0,r).$$
On the other hand, let $zin B(0,r)$, then $|z-0|<r$. This implies that
$$bigg|dfrac{z}{r}-0bigg|<1.$$
Thus, $dfrac{z}{r}in B(0,1)$ and there exists $xin B(0,1)$ such that
$$dfrac{z}{r}=xiff z=rx in rB(0,1)$$
Hence, $$ B(0,r) subseteq rB(0,1).$$
Therefore,
$$ B(0,r) = rB(0,1).$$
Question
Kindly help check if my proof is correct.
general-topology functional-analysis metric-spaces
asked Jan 18 at 11:30
Omojola MichealOmojola Micheal
1,918324
$endgroup$
If $X$ is a real normed linear space, I want to prove that $$B_{r}(0)=r B_{1}(0)$$
My proof
Let $zin rB(0,1)$, there exists $xin B(0,1)$ such that $z=rx$.
$$|z-0|=|rx-0|=r|x-0|<r$$
So, $zin B(0,r)$ and $$rB(0,1)subseteq B(0,r).$$
On the other hand, let $zin B(0,r)$, then $|z-0|<r$. This implies that
$$bigg|dfrac{z}{r}-0bigg|<1.$$
Thus, $dfrac{z}{r}in B(0,1)$ and there exists $xin B(0,1)$ such that
$$dfrac{z}{r}=xiff z=rx in rB(0,1)$$
Hence, $$ B(0,r) subseteq rB(0,1).$$
Therefore,
$$ B(0,r) = rB(0,1).$$
Question
Kindly help check if my proof is correct.
general-topology functional-analysis metric-spaces
general-topology functional-analysis metric-spaces
asked Jan 18 at 11:30
Omojola MichealOmojola Micheal
1,918324
asked Jan 18 at 11:30
Omojola MichealOmojola Micheal
1,918324
edited Jan 18 at 14:02
Omojola Micheal
asked Jan 18 at 11:30
Omojola MichealOmojola Micheal
1,918324
asked Jan 18 at 11:30
Omojola MichealOmojola Micheal
1,918324
asked Jan 18 at 11:30
Omojola MichealOmojola Micheal
1,918324
1,918324
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
The proof is totally fine. I would have written it thusly:
Suppose $r>0$.
Let $x in B_r(0)$. Then define $y=frac{1}{r}cdot x$. Then
$$|y-0| = |y|=frac{1}{r}|x|= frac{1}{r}|x-0|< frac{1}{r} cdot r = 1$$
so that $y in B_1(0)$ and as $rcdot y = r cdot (frac{1}{r}cdot x)= (r frac{1}{r}) cdot x = 1cdot x = x$, we have that $x in rB_1(0)$. So $B_r(0) subseteq rB_1(0)$.
On the other hand, if $xin rB_1(0)$ then $x=ry$ for some $y in B_1(0)$ and then
$$|x-0|=|x|=|ry| = r|y| < r cdot 1= r$$ and so $x in B_r(0)$. So the other inclusion $rB_1(0) subseteq B_r(0)$ also holds.
answered Jan 18 at 18:05
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Thanks a lot, Henno Brandsma! (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 20:23
add a comment |
$begingroup$
Your proof is fine. Everything is O.K.
answered Jan 18 at 11:45
FredFred
47k1848
$endgroup$
$begingroup$
Thanks a lot (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 15:05
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The proof is totally fine. I would have written it thusly:
Suppose $r>0$.
Let $x in B_r(0)$. Then define $y=frac{1}{r}cdot x$. Then
$$|y-0| = |y|=frac{1}{r}|x|= frac{1}{r}|x-0|< frac{1}{r} cdot r = 1$$
so that $y in B_1(0)$ and as $rcdot y = r cdot (frac{1}{r}cdot x)= (r frac{1}{r}) cdot x = 1cdot x = x$, we have that $x in rB_1(0)$. So $B_r(0) subseteq rB_1(0)$.
On the other hand, if $xin rB_1(0)$ then $x=ry$ for some $y in B_1(0)$ and then
$$|x-0|=|x|=|ry| = r|y| < r cdot 1= r$$ and so $x in B_r(0)$. So the other inclusion $rB_1(0) subseteq B_r(0)$ also holds.
answered Jan 18 at 18:05
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Thanks a lot, Henno Brandsma! (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 20:23
add a comment |
$begingroup$
The proof is totally fine. I would have written it thusly:
Suppose $r>0$.
Let $x in B_r(0)$. Then define $y=frac{1}{r}cdot x$. Then
$$|y-0| = |y|=frac{1}{r}|x|= frac{1}{r}|x-0|< frac{1}{r} cdot r = 1$$
so that $y in B_1(0)$ and as $rcdot y = r cdot (frac{1}{r}cdot x)= (r frac{1}{r}) cdot x = 1cdot x = x$, we have that $x in rB_1(0)$. So $B_r(0) subseteq rB_1(0)$.
On the other hand, if $xin rB_1(0)$ then $x=ry$ for some $y in B_1(0)$ and then
$$|x-0|=|x|=|ry| = r|y| < r cdot 1= r$$ and so $x in B_r(0)$. So the other inclusion $rB_1(0) subseteq B_r(0)$ also holds.
answered Jan 18 at 18:05
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Thanks a lot, Henno Brandsma! (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 20:23
add a comment |
$begingroup$
The proof is totally fine. I would have written it thusly:
Suppose $r>0$.
Let $x in B_r(0)$. Then define $y=frac{1}{r}cdot x$. Then
$$|y-0| = |y|=frac{1}{r}|x|= frac{1}{r}|x-0|< frac{1}{r} cdot r = 1$$
so that $y in B_1(0)$ and as $rcdot y = r cdot (frac{1}{r}cdot x)= (r frac{1}{r}) cdot x = 1cdot x = x$, we have that $x in rB_1(0)$. So $B_r(0) subseteq rB_1(0)$.
On the other hand, if $xin rB_1(0)$ then $x=ry$ for some $y in B_1(0)$ and then
$$|x-0|=|x|=|ry| = r|y| < r cdot 1= r$$ and so $x in B_r(0)$. So the other inclusion $rB_1(0) subseteq B_r(0)$ also holds.
answered Jan 18 at 18:05
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
The proof is totally fine. I would have written it thusly:
Suppose $r>0$.
Let $x in B_r(0)$. Then define $y=frac{1}{r}cdot x$. Then
$$|y-0| = |y|=frac{1}{r}|x|= frac{1}{r}|x-0|< frac{1}{r} cdot r = 1$$
so that $y in B_1(0)$ and as $rcdot y = r cdot (frac{1}{r}cdot x)= (r frac{1}{r}) cdot x = 1cdot x = x$, we have that $x in rB_1(0)$. So $B_r(0) subseteq rB_1(0)$.
On the other hand, if $xin rB_1(0)$ then $x=ry$ for some $y in B_1(0)$ and then
$$|x-0|=|x|=|ry| = r|y| < r cdot 1= r$$ and so $x in B_r(0)$. So the other inclusion $rB_1(0) subseteq B_r(0)$ also holds.
answered Jan 18 at 18:05
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 18 at 18:05
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 18 at 18:05
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 18 at 18:05
Henno BrandsmaHenno Brandsma
111k348118
111k348118
$begingroup$
Thanks a lot, Henno Brandsma! (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 20:23
add a comment |
$begingroup$
Thanks a lot, Henno Brandsma! (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 20:23
$begingroup$
Thanks a lot, Henno Brandsma! (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 20:23
$begingroup$
Thanks a lot, Henno Brandsma! (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 20:23
add a comment |
$begingroup$
Your proof is fine. Everything is O.K.
answered Jan 18 at 11:45
FredFred
47k1848
$endgroup$
$begingroup$
Thanks a lot (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 15:05
add a comment |
$begingroup$
Your proof is fine. Everything is O.K.
answered Jan 18 at 11:45
FredFred
47k1848
$endgroup$
$begingroup$
Thanks a lot (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 15:05
add a comment |
$begingroup$
Your proof is fine. Everything is O.K.
answered Jan 18 at 11:45
FredFred
47k1848
$endgroup$
Your proof is fine. Everything is O.K.
answered Jan 18 at 11:45
FredFred
47k1848
answered Jan 18 at 11:45
FredFred
47k1848
answered Jan 18 at 11:45
FredFred
47k1848
answered Jan 18 at 11:45
FredFred
47k1848
47k1848
$begingroup$
Thanks a lot (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 15:05
add a comment |
$begingroup$
Thanks a lot (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 15:05
$begingroup$
Thanks a lot (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 15:05
$begingroup$
Thanks a lot (+1)
$endgroup$
– Omojola Micheal
Jan 18 at 15:05
add a comment |
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