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Compactness of a metric space
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If a metric space $(X,d)$ is compact then for every equivalent metric $sigma$, $(X,sigma)$ is complete. This is because, for any cauchy sequence in $(X,sigma)$ has a convergent subsequence due to fact $(X,sigma)$ is a compact metric space, hence original sequence is convergent. My question is , does the converse also hold ?
That is, let $(X,d)$ be a metric space such that for every equivalent metric $sigma$, $(X,sigma)$ is complete. Does this imply $(X,d)$ is compact?
metric-spaces compactness complete-spaces
edited 1 min ago
José Carlos Santos
123k17101186
asked 2 hours ago
UserS
333
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up vote
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If a metric space $(X,d)$ is compact then for every equivalent metric $sigma$, $(X,sigma)$ is complete. This is because, for any cauchy sequence in $(X,sigma)$ has a convergent subsequence due to fact $(X,sigma)$ is a compact metric space, hence original sequence is convergent. My question is , does the converse also hold ?
That is, let $(X,d)$ be a metric space such that for every equivalent metric $sigma$, $(X,sigma)$ is complete. Does this imply $(X,d)$ is compact?
metric-spaces compactness complete-spaces
edited 1 min ago
José Carlos Santos
123k17101186
asked 2 hours ago
UserS
333
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
If a metric space $(X,d)$ is compact then for every equivalent metric $sigma$, $(X,sigma)$ is complete. This is because, for any cauchy sequence in $(X,sigma)$ has a convergent subsequence due to fact $(X,sigma)$ is a compact metric space, hence original sequence is convergent. My question is , does the converse also hold ?
That is, let $(X,d)$ be a metric space such that for every equivalent metric $sigma$, $(X,sigma)$ is complete. Does this imply $(X,d)$ is compact?
metric-spaces compactness complete-spaces
edited 1 min ago
José Carlos Santos
123k17101186
asked 2 hours ago
UserS
333
If a metric space $(X,d)$ is compact then for every equivalent metric $sigma$, $(X,sigma)$ is complete. This is because, for any cauchy sequence in $(X,sigma)$ has a convergent subsequence due to fact $(X,sigma)$ is a compact metric space, hence original sequence is convergent. My question is , does the converse also hold ?
That is, let $(X,d)$ be a metric space such that for every equivalent metric $sigma$, $(X,sigma)$ is complete. Does this imply $(X,d)$ is compact?
metric-spaces compactness complete-spaces
metric-spaces compactness complete-spaces
edited 1 min ago
José Carlos Santos
123k17101186
asked 2 hours ago
UserS
333
edited 1 min ago
José Carlos Santos
123k17101186
asked 2 hours ago
UserS
333
edited 1 min ago
José Carlos Santos
123k17101186
edited 1 min ago
José Carlos Santos
123k17101186
edited 1 min ago
José Carlos Santos
123k17101186
123k17101186
asked 2 hours ago
UserS
333
asked 2 hours ago
UserS
333
asked 2 hours ago
UserS
333
333
add a comment |Â
add a comment |Â
3 Answers
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This is true but not very easy to prove.
Suppose $X$ is not compact. Without loss of generality assume that the
original metric $d$ on $X$ is such that $d(x,y)<1$ for all $x,yin X.$ There
exists a decreasing sequence of non-empty closed sets $C_n$ whose
intersection is empty. Let $$rho (x,y)=sum_n=1^infty frac1%
2^nd_n(x,y)$$ where $$d_n(x,y)=leftvert
d(x,C_n)-d(y,C_n)rightvert +min d(x,C_n),d(y,C_n)d(x,y).$$ We
claim that $rho $ is a metric on $X$ which is equivalent to $d$ and that $%
(X,rho )$ is not complete. Note that $d_n(x,y)leq 2$ for all $x,yin X.$
If $x$ and $y$ $in C_k$ then $x$ and $y$ $in C_n$ for $1leq nleq k$
and hence $rho (x,y)leq sum_n=k+1^infty frac22^n=%
frac12^k$. Thus, the diameter of $C_k$ in $(X,rho )$ does not
exceed $frac12^k$. Once we prove that $rho $ is a metric equivalent
to $d$ it follows that $rho $ is not complete because $C_n$ is a
decreasing sequence of non-empty closed sets whose intersection is empty.
Assuming (for the time being) that $d_n$ satisfies triangle inequality it
follows easily that $rho $ is a metric: if $rho (x,y)=0$ then $%
d(x,C_n)=d(y,C_n)$ for each $n$ and $min
d(x,C_n),d(y,C_n)d(x,y)=0$ for each $n$. If $d(x,y)neq 0$ it
follows that $d(x,C_n)=d(y,C_n)=0$ for each $n$ which implies that $x$
and $y$ belong to each $C_n$ contradicting the hypothesis. Thus $rho $
is a metric. Also $rho (x_j,x)rightarrow 0$ as $jrightarrow infty $
implies $leftvert d(x_j,C_n)-d(x,C_n)rightvert rightarrow 0$ and $%
min d(x_j,C_n),d(x,C_n)d(x_j,x)rightarrow 0$ as $jrightarrow
infty $ for each $n$. There is at least one integer $k$ such that $xnotin
C_k$ and we conclude that $d(x_j,x)rightarrow 0$. Conversely, suppose $%
d(x_j,x)rightarrow 0$. Then $d_n(x_j,x)rightarrow 0$ for each $n$
and the series defining $rho $ is uniformly convergent, so $rho
(x_j,x)rightarrow 0$. It remains only to show that $d_n$ satisfies
triangle inequality for each $n$ . We have to show that $$leftvert
d(x,C_n)-d(y,C_n)rightvert$$ $$+min d(x,C_n),d(y,C_n)d(x,y)$$
$$leq leftvert d(x,C_n)-d(z,C_n)rightvert +min
d(x,C_n),d(z,C_n)d(x,z)$$ $$+leftvert d(z,C_n)-d(y,C_n)rightvert
+min d(z,C_n),d(y,C_n)d(z,y)$$ for all $x,y,z.$ Let $%
r_1=d(x,C_n),r_2=d(y,C_n),r_3=d(z,C_n)$. We consider six cases
depending on the way the numbers $r_1,r_2,r_3$ are ordered. It turns
out that the proof is easy when $r_1$ or $r_2$ is the smallest of the
three. We give the proof for the case $r_3leq r_1leq r_2$. (The case
$r_3leq r_2leq r_1$ is similar). We have to show that
$$r_2-r_1+r_1d(x,y)leq r_1-r_3+r_3d(x,z)+r_2-r_3+r_3d(z,y)$$
which says $$r_1d(x,y)leq 2r_1-2r_3+r_3d(x,z)+r_3d(z,y).$$ Since $d$
satisfies trangle inequality it suffices to show that $$%
r_1d(x,z)+r_1d(z,y)leq 2r_1-2r_3+r_3d(x,z)+r_3d(z,y).$$ But this
last inequality is equivalent to $$(r_1-r_3)[d(x,z)+d(z,y)]leq
2r_1-2r_3.$$ This is true because $d(x,z)+d(z,y)leq 1+1=2$.
answered 2 hours ago
Kavi Rama Murthy
26.9k31438
I believe I got this long ago from American Mathematical Monthly. Not my original proof.
– Kavi Rama Murthy
1 hour ago
add a comment |Â
up vote
3
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Here is an easier proof. Assume $(X,d)$ is not compact. Let $S$ be the one-point compactification of $X$. By the metrization theorems, $S$ is metrizable. Let $rho$ be such a metrization of $S$ and let $sigma$ be the restriction of $rho$ to $X$. Then $(X,sigma)$ is clearly not complete, since there exists a sequence in $X$ that converges (in $(S,rho)$) to the unique point in $Ssetminus X$. This completes the proof.
answered 19 mins ago
Ali Khezeli
935
The metrization theorem of Urysohn in the Wikipedia page is for second countable spaces.
– Kavi Rama Murthy
3 mins ago
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up vote
3
down vote
Nice question. Yes, it implies that $(X,d)$ is compact. In fact, suppose that $(X,d)$ is not compact. Then there is a sequence $(F_n)_ninmathbb N$ of closed subspaces of $X$ such that $bigcap_ninmathbb NF_n=emptyset$. For such a sequence, define the distance$$sigma(x,y)=sum_n=1^inftyfracbigllvert d (x,F_n)-d(y,F_n)bigrrvert+minbigld(x,F_n),d(y,F_n)bigrtimes d(x,y)2^n.$$It can be proved (see Ryszard Engelking's General Topology, section 4.3) that:
$sigma$ is a distance equivalent to $d$;
$(X,sigma)$ is not complete.
answered 1 hour ago
José Carlos Santos
123k17101186
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
This is true but not very easy to prove.
Suppose $X$ is not compact. Without loss of generality assume that the
original metric $d$ on $X$ is such that $d(x,y)<1$ for all $x,yin X.$ There
exists a decreasing sequence of non-empty closed sets $C_n$ whose
intersection is empty. Let $$rho (x,y)=sum_n=1^infty frac1%
2^nd_n(x,y)$$ where $$d_n(x,y)=leftvert
d(x,C_n)-d(y,C_n)rightvert +min d(x,C_n),d(y,C_n)d(x,y).$$ We
claim that $rho $ is a metric on $X$ which is equivalent to $d$ and that $%
(X,rho )$ is not complete. Note that $d_n(x,y)leq 2$ for all $x,yin X.$
If $x$ and $y$ $in C_k$ then $x$ and $y$ $in C_n$ for $1leq nleq k$
and hence $rho (x,y)leq sum_n=k+1^infty frac22^n=%
frac12^k$. Thus, the diameter of $C_k$ in $(X,rho )$ does not
exceed $frac12^k$. Once we prove that $rho $ is a metric equivalent
to $d$ it follows that $rho $ is not complete because $C_n$ is a
decreasing sequence of non-empty closed sets whose intersection is empty.
Assuming (for the time being) that $d_n$ satisfies triangle inequality it
follows easily that $rho $ is a metric: if $rho (x,y)=0$ then $%
d(x,C_n)=d(y,C_n)$ for each $n$ and $min
d(x,C_n),d(y,C_n)d(x,y)=0$ for each $n$. If $d(x,y)neq 0$ it
follows that $d(x,C_n)=d(y,C_n)=0$ for each $n$ which implies that $x$
and $y$ belong to each $C_n$ contradicting the hypothesis. Thus $rho $
is a metric. Also $rho (x_j,x)rightarrow 0$ as $jrightarrow infty $
implies $leftvert d(x_j,C_n)-d(x,C_n)rightvert rightarrow 0$ and $%
min d(x_j,C_n),d(x,C_n)d(x_j,x)rightarrow 0$ as $jrightarrow
infty $ for each $n$. There is at least one integer $k$ such that $xnotin
C_k$ and we conclude that $d(x_j,x)rightarrow 0$. Conversely, suppose $%
d(x_j,x)rightarrow 0$. Then $d_n(x_j,x)rightarrow 0$ for each $n$
and the series defining $rho $ is uniformly convergent, so $rho
(x_j,x)rightarrow 0$. It remains only to show that $d_n$ satisfies
triangle inequality for each $n$ . We have to show that $$leftvert
d(x,C_n)-d(y,C_n)rightvert$$ $$+min d(x,C_n),d(y,C_n)d(x,y)$$
$$leq leftvert d(x,C_n)-d(z,C_n)rightvert +min
d(x,C_n),d(z,C_n)d(x,z)$$ $$+leftvert d(z,C_n)-d(y,C_n)rightvert
+min d(z,C_n),d(y,C_n)d(z,y)$$ for all $x,y,z.$ Let $%
r_1=d(x,C_n),r_2=d(y,C_n),r_3=d(z,C_n)$. We consider six cases
depending on the way the numbers $r_1,r_2,r_3$ are ordered. It turns
out that the proof is easy when $r_1$ or $r_2$ is the smallest of the
three. We give the proof for the case $r_3leq r_1leq r_2$. (The case
$r_3leq r_2leq r_1$ is similar). We have to show that
$$r_2-r_1+r_1d(x,y)leq r_1-r_3+r_3d(x,z)+r_2-r_3+r_3d(z,y)$$
which says $$r_1d(x,y)leq 2r_1-2r_3+r_3d(x,z)+r_3d(z,y).$$ Since $d$
satisfies trangle inequality it suffices to show that $$%
r_1d(x,z)+r_1d(z,y)leq 2r_1-2r_3+r_3d(x,z)+r_3d(z,y).$$ But this
last inequality is equivalent to $$(r_1-r_3)[d(x,z)+d(z,y)]leq
2r_1-2r_3.$$ This is true because $d(x,z)+d(z,y)leq 1+1=2$.
answered 2 hours ago
Kavi Rama Murthy
26.9k31438
I believe I got this long ago from American Mathematical Monthly. Not my original proof.
– Kavi Rama Murthy
1 hour ago
add a comment |Â
up vote
4
down vote
This is true but not very easy to prove.
Suppose $X$ is not compact. Without loss of generality assume that the
original metric $d$ on $X$ is such that $d(x,y)<1$ for all $x,yin X.$ There
exists a decreasing sequence of non-empty closed sets $C_n$ whose
intersection is empty. Let $$rho (x,y)=sum_n=1^infty frac1%
2^nd_n(x,y)$$ where $$d_n(x,y)=leftvert
d(x,C_n)-d(y,C_n)rightvert +min d(x,C_n),d(y,C_n)d(x,y).$$ We
claim that $rho $ is a metric on $X$ which is equivalent to $d$ and that $%
(X,rho )$ is not complete. Note that $d_n(x,y)leq 2$ for all $x,yin X.$
If $x$ and $y$ $in C_k$ then $x$ and $y$ $in C_n$ for $1leq nleq k$
and hence $rho (x,y)leq sum_n=k+1^infty frac22^n=%
frac12^k$. Thus, the diameter of $C_k$ in $(X,rho )$ does not
exceed $frac12^k$. Once we prove that $rho $ is a metric equivalent
to $d$ it follows that $rho $ is not complete because $C_n$ is a
decreasing sequence of non-empty closed sets whose intersection is empty.
Assuming (for the time being) that $d_n$ satisfies triangle inequality it
follows easily that $rho $ is a metric: if $rho (x,y)=0$ then $%
d(x,C_n)=d(y,C_n)$ for each $n$ and $min
d(x,C_n),d(y,C_n)d(x,y)=0$ for each $n$. If $d(x,y)neq 0$ it
follows that $d(x,C_n)=d(y,C_n)=0$ for each $n$ which implies that $x$
and $y$ belong to each $C_n$ contradicting the hypothesis. Thus $rho $
is a metric. Also $rho (x_j,x)rightarrow 0$ as $jrightarrow infty $
implies $leftvert d(x_j,C_n)-d(x,C_n)rightvert rightarrow 0$ and $%
min d(x_j,C_n),d(x,C_n)d(x_j,x)rightarrow 0$ as $jrightarrow
infty $ for each $n$. There is at least one integer $k$ such that $xnotin
C_k$ and we conclude that $d(x_j,x)rightarrow 0$. Conversely, suppose $%
d(x_j,x)rightarrow 0$. Then $d_n(x_j,x)rightarrow 0$ for each $n$
and the series defining $rho $ is uniformly convergent, so $rho
(x_j,x)rightarrow 0$. It remains only to show that $d_n$ satisfies
triangle inequality for each $n$ . We have to show that $$leftvert
d(x,C_n)-d(y,C_n)rightvert$$ $$+min d(x,C_n),d(y,C_n)d(x,y)$$
$$leq leftvert d(x,C_n)-d(z,C_n)rightvert +min
d(x,C_n),d(z,C_n)d(x,z)$$ $$+leftvert d(z,C_n)-d(y,C_n)rightvert
+min d(z,C_n),d(y,C_n)d(z,y)$$ for all $x,y,z.$ Let $%
r_1=d(x,C_n),r_2=d(y,C_n),r_3=d(z,C_n)$. We consider six cases
depending on the way the numbers $r_1,r_2,r_3$ are ordered. It turns
out that the proof is easy when $r_1$ or $r_2$ is the smallest of the
three. We give the proof for the case $r_3leq r_1leq r_2$. (The case
$r_3leq r_2leq r_1$ is similar). We have to show that
$$r_2-r_1+r_1d(x,y)leq r_1-r_3+r_3d(x,z)+r_2-r_3+r_3d(z,y)$$
which says $$r_1d(x,y)leq 2r_1-2r_3+r_3d(x,z)+r_3d(z,y).$$ Since $d$
satisfies trangle inequality it suffices to show that $$%
r_1d(x,z)+r_1d(z,y)leq 2r_1-2r_3+r_3d(x,z)+r_3d(z,y).$$ But this
last inequality is equivalent to $$(r_1-r_3)[d(x,z)+d(z,y)]leq
2r_1-2r_3.$$ This is true because $d(x,z)+d(z,y)leq 1+1=2$.
answered 2 hours ago
Kavi Rama Murthy
26.9k31438
I believe I got this long ago from American Mathematical Monthly. Not my original proof.
– Kavi Rama Murthy
1 hour ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
This is true but not very easy to prove.
Suppose $X$ is not compact. Without loss of generality assume that the
original metric $d$ on $X$ is such that $d(x,y)<1$ for all $x,yin X.$ There
exists a decreasing sequence of non-empty closed sets $C_n$ whose
intersection is empty. Let $$rho (x,y)=sum_n=1^infty frac1%
2^nd_n(x,y)$$ where $$d_n(x,y)=leftvert
d(x,C_n)-d(y,C_n)rightvert +min d(x,C_n),d(y,C_n)d(x,y).$$ We
claim that $rho $ is a metric on $X$ which is equivalent to $d$ and that $%
(X,rho )$ is not complete. Note that $d_n(x,y)leq 2$ for all $x,yin X.$
If $x$ and $y$ $in C_k$ then $x$ and $y$ $in C_n$ for $1leq nleq k$
and hence $rho (x,y)leq sum_n=k+1^infty frac22^n=%
frac12^k$. Thus, the diameter of $C_k$ in $(X,rho )$ does not
exceed $frac12^k$. Once we prove that $rho $ is a metric equivalent
to $d$ it follows that $rho $ is not complete because $C_n$ is a
decreasing sequence of non-empty closed sets whose intersection is empty.
Assuming (for the time being) that $d_n$ satisfies triangle inequality it
follows easily that $rho $ is a metric: if $rho (x,y)=0$ then $%
d(x,C_n)=d(y,C_n)$ for each $n$ and $min
d(x,C_n),d(y,C_n)d(x,y)=0$ for each $n$. If $d(x,y)neq 0$ it
follows that $d(x,C_n)=d(y,C_n)=0$ for each $n$ which implies that $x$
and $y$ belong to each $C_n$ contradicting the hypothesis. Thus $rho $
is a metric. Also $rho (x_j,x)rightarrow 0$ as $jrightarrow infty $
implies $leftvert d(x_j,C_n)-d(x,C_n)rightvert rightarrow 0$ and $%
min d(x_j,C_n),d(x,C_n)d(x_j,x)rightarrow 0$ as $jrightarrow
infty $ for each $n$. There is at least one integer $k$ such that $xnotin
C_k$ and we conclude that $d(x_j,x)rightarrow 0$. Conversely, suppose $%
d(x_j,x)rightarrow 0$. Then $d_n(x_j,x)rightarrow 0$ for each $n$
and the series defining $rho $ is uniformly convergent, so $rho
(x_j,x)rightarrow 0$. It remains only to show that $d_n$ satisfies
triangle inequality for each $n$ . We have to show that $$leftvert
d(x,C_n)-d(y,C_n)rightvert$$ $$+min d(x,C_n),d(y,C_n)d(x,y)$$
$$leq leftvert d(x,C_n)-d(z,C_n)rightvert +min
d(x,C_n),d(z,C_n)d(x,z)$$ $$+leftvert d(z,C_n)-d(y,C_n)rightvert
+min d(z,C_n),d(y,C_n)d(z,y)$$ for all $x,y,z.$ Let $%
r_1=d(x,C_n),r_2=d(y,C_n),r_3=d(z,C_n)$. We consider six cases
depending on the way the numbers $r_1,r_2,r_3$ are ordered. It turns
out that the proof is easy when $r_1$ or $r_2$ is the smallest of the
three. We give the proof for the case $r_3leq r_1leq r_2$. (The case
$r_3leq r_2leq r_1$ is similar). We have to show that
$$r_2-r_1+r_1d(x,y)leq r_1-r_3+r_3d(x,z)+r_2-r_3+r_3d(z,y)$$
which says $$r_1d(x,y)leq 2r_1-2r_3+r_3d(x,z)+r_3d(z,y).$$ Since $d$
satisfies trangle inequality it suffices to show that $$%
r_1d(x,z)+r_1d(z,y)leq 2r_1-2r_3+r_3d(x,z)+r_3d(z,y).$$ But this
last inequality is equivalent to $$(r_1-r_3)[d(x,z)+d(z,y)]leq
2r_1-2r_3.$$ This is true because $d(x,z)+d(z,y)leq 1+1=2$.
answered 2 hours ago
Kavi Rama Murthy
26.9k31438
This is true but not very easy to prove.
Suppose $X$ is not compact. Without loss of generality assume that the
original metric $d$ on $X$ is such that $d(x,y)<1$ for all $x,yin X.$ There
exists a decreasing sequence of non-empty closed sets $C_n$ whose
intersection is empty. Let $$rho (x,y)=sum_n=1^infty frac1%
2^nd_n(x,y)$$ where $$d_n(x,y)=leftvert
d(x,C_n)-d(y,C_n)rightvert +min d(x,C_n),d(y,C_n)d(x,y).$$ We
claim that $rho $ is a metric on $X$ which is equivalent to $d$ and that $%
(X,rho )$ is not complete. Note that $d_n(x,y)leq 2$ for all $x,yin X.$
If $x$ and $y$ $in C_k$ then $x$ and $y$ $in C_n$ for $1leq nleq k$
and hence $rho (x,y)leq sum_n=k+1^infty frac22^n=%
frac12^k$. Thus, the diameter of $C_k$ in $(X,rho )$ does not
exceed $frac12^k$. Once we prove that $rho $ is a metric equivalent
to $d$ it follows that $rho $ is not complete because $C_n$ is a
decreasing sequence of non-empty closed sets whose intersection is empty.
Assuming (for the time being) that $d_n$ satisfies triangle inequality it
follows easily that $rho $ is a metric: if $rho (x,y)=0$ then $%
d(x,C_n)=d(y,C_n)$ for each $n$ and $min
d(x,C_n),d(y,C_n)d(x,y)=0$ for each $n$. If $d(x,y)neq 0$ it
follows that $d(x,C_n)=d(y,C_n)=0$ for each $n$ which implies that $x$
and $y$ belong to each $C_n$ contradicting the hypothesis. Thus $rho $
is a metric. Also $rho (x_j,x)rightarrow 0$ as $jrightarrow infty $
implies $leftvert d(x_j,C_n)-d(x,C_n)rightvert rightarrow 0$ and $%
min d(x_j,C_n),d(x,C_n)d(x_j,x)rightarrow 0$ as $jrightarrow
infty $ for each $n$. There is at least one integer $k$ such that $xnotin
C_k$ and we conclude that $d(x_j,x)rightarrow 0$. Conversely, suppose $%
d(x_j,x)rightarrow 0$. Then $d_n(x_j,x)rightarrow 0$ for each $n$
and the series defining $rho $ is uniformly convergent, so $rho
(x_j,x)rightarrow 0$. It remains only to show that $d_n$ satisfies
triangle inequality for each $n$ . We have to show that $$leftvert
d(x,C_n)-d(y,C_n)rightvert$$ $$+min d(x,C_n),d(y,C_n)d(x,y)$$
$$leq leftvert d(x,C_n)-d(z,C_n)rightvert +min
d(x,C_n),d(z,C_n)d(x,z)$$ $$+leftvert d(z,C_n)-d(y,C_n)rightvert
+min d(z,C_n),d(y,C_n)d(z,y)$$ for all $x,y,z.$ Let $%
r_1=d(x,C_n),r_2=d(y,C_n),r_3=d(z,C_n)$. We consider six cases
depending on the way the numbers $r_1,r_2,r_3$ are ordered. It turns
out that the proof is easy when $r_1$ or $r_2$ is the smallest of the
three. We give the proof for the case $r_3leq r_1leq r_2$. (The case
$r_3leq r_2leq r_1$ is similar). We have to show that
$$r_2-r_1+r_1d(x,y)leq r_1-r_3+r_3d(x,z)+r_2-r_3+r_3d(z,y)$$
which says $$r_1d(x,y)leq 2r_1-2r_3+r_3d(x,z)+r_3d(z,y).$$ Since $d$
satisfies trangle inequality it suffices to show that $$%
r_1d(x,z)+r_1d(z,y)leq 2r_1-2r_3+r_3d(x,z)+r_3d(z,y).$$ But this
last inequality is equivalent to $$(r_1-r_3)[d(x,z)+d(z,y)]leq
2r_1-2r_3.$$ This is true because $d(x,z)+d(z,y)leq 1+1=2$.
answered 2 hours ago
Kavi Rama Murthy
26.9k31438
edited 1 hour ago
answered 2 hours ago
Kavi Rama Murthy
26.9k31438
answered 2 hours ago
Kavi Rama Murthy
26.9k31438
answered 2 hours ago
Kavi Rama Murthy
26.9k31438
26.9k31438
I believe I got this long ago from American Mathematical Monthly. Not my original proof.
– Kavi Rama Murthy
1 hour ago
add a comment |Â
I believe I got this long ago from American Mathematical Monthly. Not my original proof.
– Kavi Rama Murthy
1 hour ago
I believe I got this long ago from American Mathematical Monthly. Not my original proof.
– Kavi Rama Murthy
1 hour ago
I believe I got this long ago from American Mathematical Monthly. Not my original proof.
– Kavi Rama Murthy
1 hour ago
add a comment |Â
up vote
3
down vote
Here is an easier proof. Assume $(X,d)$ is not compact. Let $S$ be the one-point compactification of $X$. By the metrization theorems, $S$ is metrizable. Let $rho$ be such a metrization of $S$ and let $sigma$ be the restriction of $rho$ to $X$. Then $(X,sigma)$ is clearly not complete, since there exists a sequence in $X$ that converges (in $(S,rho)$) to the unique point in $Ssetminus X$. This completes the proof.
answered 19 mins ago
Ali Khezeli
935
The metrization theorem of Urysohn in the Wikipedia page is for second countable spaces.
– Kavi Rama Murthy
3 mins ago
add a comment |Â
up vote
3
down vote
Here is an easier proof. Assume $(X,d)$ is not compact. Let $S$ be the one-point compactification of $X$. By the metrization theorems, $S$ is metrizable. Let $rho$ be such a metrization of $S$ and let $sigma$ be the restriction of $rho$ to $X$. Then $(X,sigma)$ is clearly not complete, since there exists a sequence in $X$ that converges (in $(S,rho)$) to the unique point in $Ssetminus X$. This completes the proof.
answered 19 mins ago
Ali Khezeli
935
The metrization theorem of Urysohn in the Wikipedia page is for second countable spaces.
– Kavi Rama Murthy
3 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Here is an easier proof. Assume $(X,d)$ is not compact. Let $S$ be the one-point compactification of $X$. By the metrization theorems, $S$ is metrizable. Let $rho$ be such a metrization of $S$ and let $sigma$ be the restriction of $rho$ to $X$. Then $(X,sigma)$ is clearly not complete, since there exists a sequence in $X$ that converges (in $(S,rho)$) to the unique point in $Ssetminus X$. This completes the proof.
answered 19 mins ago
Ali Khezeli
935
Here is an easier proof. Assume $(X,d)$ is not compact. Let $S$ be the one-point compactification of $X$. By the metrization theorems, $S$ is metrizable. Let $rho$ be such a metrization of $S$ and let $sigma$ be the restriction of $rho$ to $X$. Then $(X,sigma)$ is clearly not complete, since there exists a sequence in $X$ that converges (in $(S,rho)$) to the unique point in $Ssetminus X$. This completes the proof.
answered 19 mins ago
Ali Khezeli
935
edited 9 mins ago
answered 19 mins ago
Ali Khezeli
935
answered 19 mins ago
Ali Khezeli
935
answered 19 mins ago
Ali Khezeli
935
935
The metrization theorem of Urysohn in the Wikipedia page is for second countable spaces.
– Kavi Rama Murthy
3 mins ago
add a comment |Â
The metrization theorem of Urysohn in the Wikipedia page is for second countable spaces.
– Kavi Rama Murthy
3 mins ago
The metrization theorem of Urysohn in the Wikipedia page is for second countable spaces.
– Kavi Rama Murthy
3 mins ago
The metrization theorem of Urysohn in the Wikipedia page is for second countable spaces.
– Kavi Rama Murthy
3 mins ago
add a comment |Â
up vote
3
down vote
Nice question. Yes, it implies that $(X,d)$ is compact. In fact, suppose that $(X,d)$ is not compact. Then there is a sequence $(F_n)_ninmathbb N$ of closed subspaces of $X$ such that $bigcap_ninmathbb NF_n=emptyset$. For such a sequence, define the distance$$sigma(x,y)=sum_n=1^inftyfracbigllvert d (x,F_n)-d(y,F_n)bigrrvert+minbigld(x,F_n),d(y,F_n)bigrtimes d(x,y)2^n.$$It can be proved (see Ryszard Engelking's General Topology, section 4.3) that:
$sigma$ is a distance equivalent to $d$;
$(X,sigma)$ is not complete.
answered 1 hour ago
José Carlos Santos
123k17101186
add a comment |Â
up vote
3
down vote
Nice question. Yes, it implies that $(X,d)$ is compact. In fact, suppose that $(X,d)$ is not compact. Then there is a sequence $(F_n)_ninmathbb N$ of closed subspaces of $X$ such that $bigcap_ninmathbb NF_n=emptyset$. For such a sequence, define the distance$$sigma(x,y)=sum_n=1^inftyfracbigllvert d (x,F_n)-d(y,F_n)bigrrvert+minbigld(x,F_n),d(y,F_n)bigrtimes d(x,y)2^n.$$It can be proved (see Ryszard Engelking's General Topology, section 4.3) that:
$sigma$ is a distance equivalent to $d$;
$(X,sigma)$ is not complete.
answered 1 hour ago
José Carlos Santos
123k17101186
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Nice question. Yes, it implies that $(X,d)$ is compact. In fact, suppose that $(X,d)$ is not compact. Then there is a sequence $(F_n)_ninmathbb N$ of closed subspaces of $X$ such that $bigcap_ninmathbb NF_n=emptyset$. For such a sequence, define the distance$$sigma(x,y)=sum_n=1^inftyfracbigllvert d (x,F_n)-d(y,F_n)bigrrvert+minbigld(x,F_n),d(y,F_n)bigrtimes d(x,y)2^n.$$It can be proved (see Ryszard Engelking's General Topology, section 4.3) that:
$sigma$ is a distance equivalent to $d$;
$(X,sigma)$ is not complete.
answered 1 hour ago
José Carlos Santos
123k17101186
Nice question. Yes, it implies that $(X,d)$ is compact. In fact, suppose that $(X,d)$ is not compact. Then there is a sequence $(F_n)_ninmathbb N$ of closed subspaces of $X$ such that $bigcap_ninmathbb NF_n=emptyset$. For such a sequence, define the distance$$sigma(x,y)=sum_n=1^inftyfracbigllvert d (x,F_n)-d(y,F_n)bigrrvert+minbigld(x,F_n),d(y,F_n)bigrtimes d(x,y)2^n.$$It can be proved (see Ryszard Engelking's General Topology, section 4.3) that:
$sigma$ is a distance equivalent to $d$;
$(X,sigma)$ is not complete.
answered 1 hour ago
José Carlos Santos
123k17101186
edited 1 min ago
answered 1 hour ago
José Carlos Santos
123k17101186
answered 1 hour ago
José Carlos Santos
123k17101186
answered 1 hour ago
José Carlos Santos
123k17101186
123k17101186
add a comment |Â
add a comment |Â
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