Prove the map has a fixed point
up vote
18
down vote
favorite
Assume $K$ is a compact metric space with metric $rho$ and $A$ is a map from $K$ to $K$ such that $rho (Ax,Ay) < rho(x,y)$ for $xneq y$. Prove A have a unique fixed point in $K$.
The uniqueness is easy. My problem is to show that there a exist fixed point. $K$ is compact, so every sequence has convergent subsequence. Construct a sequence ${x_n}$ by $x_{n+1}=Ax_{n}$,${x_n}$ has a convergent subsequence ${ x_{n_k}}$, but how to show there is a fixed point using $rho (Ax,Ay) < rho(x,y)$?
metric-spaces compactness fixed-point-theorems
edited Sep 29 '15 at 6:37
Martin Sleziak
44.4k7115267
asked Mar 10 '12 at 13:00
Jiangnan Yu
493415
add a comment |
up vote
18
down vote
favorite
Assume $K$ is a compact metric space with metric $rho$ and $A$ is a map from $K$ to $K$ such that $rho (Ax,Ay) < rho(x,y)$ for $xneq y$. Prove A have a unique fixed point in $K$.
The uniqueness is easy. My problem is to show that there a exist fixed point. $K$ is compact, so every sequence has convergent subsequence. Construct a sequence ${x_n}$ by $x_{n+1}=Ax_{n}$,${x_n}$ has a convergent subsequence ${ x_{n_k}}$, but how to show there is a fixed point using $rho (Ax,Ay) < rho(x,y)$?
metric-spaces compactness fixed-point-theorems
edited Sep 29 '15 at 6:37
Martin Sleziak
44.4k7115267
asked Mar 10 '12 at 13:00
Jiangnan Yu
493415
1
(1) I think you need to assume $K$ is complete. (2) You have a convergent subsequence; the only thing you can do now is examine the behavior of its limit ...
– Neal
Mar 10 '12 at 13:09
6
@Neal: A metric space is compact iff it is complete and totally bounded, so completeness comes for free with compactness.
– Brian M. Scott
Mar 10 '12 at 13:17
Oh, I totally missed "compact" in the question. My bad.
– Neal
Mar 11 '12 at 0:24
add a comment |
up vote
18
down vote
favorite
up vote
18
down vote
favorite
Assume $K$ is a compact metric space with metric $rho$ and $A$ is a map from $K$ to $K$ such that $rho (Ax,Ay) < rho(x,y)$ for $xneq y$. Prove A have a unique fixed point in $K$.
The uniqueness is easy. My problem is to show that there a exist fixed point. $K$ is compact, so every sequence has convergent subsequence. Construct a sequence ${x_n}$ by $x_{n+1}=Ax_{n}$,${x_n}$ has a convergent subsequence ${ x_{n_k}}$, but how to show there is a fixed point using $rho (Ax,Ay) < rho(x,y)$?
metric-spaces compactness fixed-point-theorems
edited Sep 29 '15 at 6:37
Martin Sleziak
44.4k7115267
asked Mar 10 '12 at 13:00
Jiangnan Yu
493415
Assume $K$ is a compact metric space with metric $rho$ and $A$ is a map from $K$ to $K$ such that $rho (Ax,Ay) < rho(x,y)$ for $xneq y$. Prove A have a unique fixed point in $K$.
The uniqueness is easy. My problem is to show that there a exist fixed point. $K$ is compact, so every sequence has convergent subsequence. Construct a sequence ${x_n}$ by $x_{n+1}=Ax_{n}$,${x_n}$ has a convergent subsequence ${ x_{n_k}}$, but how to show there is a fixed point using $rho (Ax,Ay) < rho(x,y)$?
metric-spaces compactness fixed-point-theorems
metric-spaces compactness fixed-point-theorems
edited Sep 29 '15 at 6:37
Martin Sleziak
44.4k7115267
asked Mar 10 '12 at 13:00
Jiangnan Yu
493415
edited Sep 29 '15 at 6:37
Martin Sleziak
44.4k7115267
asked Mar 10 '12 at 13:00
Jiangnan Yu
493415
edited Sep 29 '15 at 6:37
Martin Sleziak
44.4k7115267
edited Sep 29 '15 at 6:37
Martin Sleziak
44.4k7115267
edited Sep 29 '15 at 6:37
Martin Sleziak
44.4k7115267
44.4k7115267
asked Mar 10 '12 at 13:00
Jiangnan Yu
493415
asked Mar 10 '12 at 13:00
Jiangnan Yu
493415
asked Mar 10 '12 at 13:00
Jiangnan Yu
493415
493415
1
(1) I think you need to assume $K$ is complete. (2) You have a convergent subsequence; the only thing you can do now is examine the behavior of its limit ...
– Neal
Mar 10 '12 at 13:09
6
@Neal: A metric space is compact iff it is complete and totally bounded, so completeness comes for free with compactness.
– Brian M. Scott
Mar 10 '12 at 13:17
Oh, I totally missed "compact" in the question. My bad.
– Neal
Mar 11 '12 at 0:24
add a comment |
1
(1) I think you need to assume $K$ is complete. (2) You have a convergent subsequence; the only thing you can do now is examine the behavior of its limit ...
– Neal
Mar 10 '12 at 13:09
6
@Neal: A metric space is compact iff it is complete and totally bounded, so completeness comes for free with compactness.
– Brian M. Scott
Mar 10 '12 at 13:17
Oh, I totally missed "compact" in the question. My bad.
– Neal
Mar 11 '12 at 0:24
1
1
(1) I think you need to assume $K$ is complete. (2) You have a convergent subsequence; the only thing you can do now is examine the behavior of its limit ...
– Neal
Mar 10 '12 at 13:09
(1) I think you need to assume $K$ is complete. (2) You have a convergent subsequence; the only thing you can do now is examine the behavior of its limit ...
– Neal
Mar 10 '12 at 13:09
6
6
@Neal: A metric space is compact iff it is complete and totally bounded, so completeness comes for free with compactness.
– Brian M. Scott
Mar 10 '12 at 13:17
@Neal: A metric space is compact iff it is complete and totally bounded, so completeness comes for free with compactness.
– Brian M. Scott
Mar 10 '12 at 13:17
Oh, I totally missed "compact" in the question. My bad.
– Neal
Mar 11 '12 at 0:24
Oh, I totally missed "compact" in the question. My bad.
– Neal
Mar 11 '12 at 0:24
add a comment |
2 Answers
2
active
oldest
votes
up vote
20
down vote
accepted
Define $f(x):=rho(x,A(x))$; it's a continuous map. (Note $$rho(x,Ax)lerho(x,y)+rho(y,Ay)+rho(Ay,Ax)quadforall x, yin K$$ or $$rho(x,Ax)-rho(y,Ay)lerho(x,y)+rho(Ax,Ay).$$ Reversing the roles of $x,y$ to get $$left|rho(x,Ax)-rho(y,Ay)right|lerho(x,y)+rho(Ax,Ay)<2delta quad text{ whenever }rho(x,y)<delta.$$ That is, $f$ is actually uniformly continuous.)
Let $alpha:=inf_{xin K}f(x)$, then we can find $x_0in K$ such that $alpha=f(x_0)$, since $K$ is compact. If $alpha>0$, then $x_0neq Ax_0$ and $rho(A(Ax_0),Ax_0)<rho(Ax_0,x_0)=alpha$, which is a contradiction. So $alpha=0$ and $x_0$ is a fixed point. The assumption on $A$ makes it unique.
Note that completeness wouldn't be enough in this case, for example consider $mathbb R$ with the usual metric, and $A(x):=sqrt{x^2+1}$. It's the major difference between $rho(Ax,Ay)<rho(x,y)$ for $xneq y$ and the existence of $0<c<1$ such that for all $x,y,$: $rho(Ax,Ay)leq crho(x,y)$.
edited Oct 15 '14 at 17:46
Fang Jing
7461418
answered Mar 10 '12 at 13:15
Davide Giraudo
123k16149253
Nice proof!Thank you! :))
– Jiangnan Yu
Mar 10 '12 at 13:53
How do we show that f(x):=ρ(x,A(x)) is indeed continuous?
– Jacques
Apr 2 '12 at 9:22
3
@Jacques: $delta: x mapsto (x,x)$ is continuous, $A$ is continuous, so $g:(x,y) mapsto (x,A(y))$ is continuous, and $d:(x,y) mapsto d(x,y)$ is continuous, so $f(x) = (dcirc g circ delta)(x)$ is a composition of continuous maps, hence it is continuous. Alternatively, use the triangle inequality and the reverse triangle inequality a few times.
– t.b.
Apr 2 '12 at 9:52
Can someone clarify about uniqueness?
– Niebla
Nov 9 '15 at 2:57
1
@Niebla In general if we have $rho(A(x), A(y))<rho(x,y)$ - note that the inequality is strict - $A$ can only have one fixed point. Let $a, b$ be two fixed points, then $rho(A(a), B(b))<rho(a, b)$, which is a contradiction since both sides of this strict inequality are equal.
– Jack M
Dec 27 '15 at 16:53
|
show 1 more comment
up vote
2
down vote
I don't have enough reputation to post a comment to reply to @андрэ 's question regarding where in the proof it is used that $f$ is a continuous function, so I'll post my answer here:
Since we are told that $K$ is a compact set. $f:Krightarrow K$ being continuous implies that the $mathrm{im}(f) = f(K)$ is also a compact set. We also know that compact sets are closed and bounded, which implies the existence of $inf_{xin K} f(x)$.
If it is possible to show that $f(K) subseteq K$ is a closed set, then it is necessarily compact as well:
A subset of a compact set is compact?
However, I am not aware of how you would do this in this case without relying on continuity of $f$.
answered Oct 21 at 8:49
Matthew O'Brien
415
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
20
down vote
accepted
Define $f(x):=rho(x,A(x))$; it's a continuous map. (Note $$rho(x,Ax)lerho(x,y)+rho(y,Ay)+rho(Ay,Ax)quadforall x, yin K$$ or $$rho(x,Ax)-rho(y,Ay)lerho(x,y)+rho(Ax,Ay).$$ Reversing the roles of $x,y$ to get $$left|rho(x,Ax)-rho(y,Ay)right|lerho(x,y)+rho(Ax,Ay)<2delta quad text{ whenever }rho(x,y)<delta.$$ That is, $f$ is actually uniformly continuous.)
Let $alpha:=inf_{xin K}f(x)$, then we can find $x_0in K$ such that $alpha=f(x_0)$, since $K$ is compact. If $alpha>0$, then $x_0neq Ax_0$ and $rho(A(Ax_0),Ax_0)<rho(Ax_0,x_0)=alpha$, which is a contradiction. So $alpha=0$ and $x_0$ is a fixed point. The assumption on $A$ makes it unique.
Note that completeness wouldn't be enough in this case, for example consider $mathbb R$ with the usual metric, and $A(x):=sqrt{x^2+1}$. It's the major difference between $rho(Ax,Ay)<rho(x,y)$ for $xneq y$ and the existence of $0<c<1$ such that for all $x,y,$: $rho(Ax,Ay)leq crho(x,y)$.
edited Oct 15 '14 at 17:46
Fang Jing
7461418
answered Mar 10 '12 at 13:15
Davide Giraudo
123k16149253
Nice proof!Thank you! :))
– Jiangnan Yu
Mar 10 '12 at 13:53
How do we show that f(x):=ρ(x,A(x)) is indeed continuous?
– Jacques
Apr 2 '12 at 9:22
3
@Jacques: $delta: x mapsto (x,x)$ is continuous, $A$ is continuous, so $g:(x,y) mapsto (x,A(y))$ is continuous, and $d:(x,y) mapsto d(x,y)$ is continuous, so $f(x) = (dcirc g circ delta)(x)$ is a composition of continuous maps, hence it is continuous. Alternatively, use the triangle inequality and the reverse triangle inequality a few times.
– t.b.
Apr 2 '12 at 9:52
Can someone clarify about uniqueness?
– Niebla
Nov 9 '15 at 2:57
1
@Niebla In general if we have $rho(A(x), A(y))<rho(x,y)$ - note that the inequality is strict - $A$ can only have one fixed point. Let $a, b$ be two fixed points, then $rho(A(a), B(b))<rho(a, b)$, which is a contradiction since both sides of this strict inequality are equal.
– Jack M
Dec 27 '15 at 16:53
|
show 1 more comment
up vote
20
down vote
accepted
Define $f(x):=rho(x,A(x))$; it's a continuous map. (Note $$rho(x,Ax)lerho(x,y)+rho(y,Ay)+rho(Ay,Ax)quadforall x, yin K$$ or $$rho(x,Ax)-rho(y,Ay)lerho(x,y)+rho(Ax,Ay).$$ Reversing the roles of $x,y$ to get $$left|rho(x,Ax)-rho(y,Ay)right|lerho(x,y)+rho(Ax,Ay)<2delta quad text{ whenever }rho(x,y)<delta.$$ That is, $f$ is actually uniformly continuous.)
Let $alpha:=inf_{xin K}f(x)$, then we can find $x_0in K$ such that $alpha=f(x_0)$, since $K$ is compact. If $alpha>0$, then $x_0neq Ax_0$ and $rho(A(Ax_0),Ax_0)<rho(Ax_0,x_0)=alpha$, which is a contradiction. So $alpha=0$ and $x_0$ is a fixed point. The assumption on $A$ makes it unique.
Note that completeness wouldn't be enough in this case, for example consider $mathbb R$ with the usual metric, and $A(x):=sqrt{x^2+1}$. It's the major difference between $rho(Ax,Ay)<rho(x,y)$ for $xneq y$ and the existence of $0<c<1$ such that for all $x,y,$: $rho(Ax,Ay)leq crho(x,y)$.
edited Oct 15 '14 at 17:46
Fang Jing
7461418
answered Mar 10 '12 at 13:15
Davide Giraudo
123k16149253
Nice proof!Thank you! :))
– Jiangnan Yu
Mar 10 '12 at 13:53
How do we show that f(x):=ρ(x,A(x)) is indeed continuous?
– Jacques
Apr 2 '12 at 9:22
3
@Jacques: $delta: x mapsto (x,x)$ is continuous, $A$ is continuous, so $g:(x,y) mapsto (x,A(y))$ is continuous, and $d:(x,y) mapsto d(x,y)$ is continuous, so $f(x) = (dcirc g circ delta)(x)$ is a composition of continuous maps, hence it is continuous. Alternatively, use the triangle inequality and the reverse triangle inequality a few times.
– t.b.
Apr 2 '12 at 9:52
Can someone clarify about uniqueness?
– Niebla
Nov 9 '15 at 2:57
1
@Niebla In general if we have $rho(A(x), A(y))<rho(x,y)$ - note that the inequality is strict - $A$ can only have one fixed point. Let $a, b$ be two fixed points, then $rho(A(a), B(b))<rho(a, b)$, which is a contradiction since both sides of this strict inequality are equal.
– Jack M
Dec 27 '15 at 16:53
|
show 1 more comment
up vote
20
down vote
accepted
up vote
20
down vote
accepted
Define $f(x):=rho(x,A(x))$; it's a continuous map. (Note $$rho(x,Ax)lerho(x,y)+rho(y,Ay)+rho(Ay,Ax)quadforall x, yin K$$ or $$rho(x,Ax)-rho(y,Ay)lerho(x,y)+rho(Ax,Ay).$$ Reversing the roles of $x,y$ to get $$left|rho(x,Ax)-rho(y,Ay)right|lerho(x,y)+rho(Ax,Ay)<2delta quad text{ whenever }rho(x,y)<delta.$$ That is, $f$ is actually uniformly continuous.)
Let $alpha:=inf_{xin K}f(x)$, then we can find $x_0in K$ such that $alpha=f(x_0)$, since $K$ is compact. If $alpha>0$, then $x_0neq Ax_0$ and $rho(A(Ax_0),Ax_0)<rho(Ax_0,x_0)=alpha$, which is a contradiction. So $alpha=0$ and $x_0$ is a fixed point. The assumption on $A$ makes it unique.
Note that completeness wouldn't be enough in this case, for example consider $mathbb R$ with the usual metric, and $A(x):=sqrt{x^2+1}$. It's the major difference between $rho(Ax,Ay)<rho(x,y)$ for $xneq y$ and the existence of $0<c<1$ such that for all $x,y,$: $rho(Ax,Ay)leq crho(x,y)$.
edited Oct 15 '14 at 17:46
Fang Jing
7461418
answered Mar 10 '12 at 13:15
Davide Giraudo
123k16149253
Define $f(x):=rho(x,A(x))$; it's a continuous map. (Note $$rho(x,Ax)lerho(x,y)+rho(y,Ay)+rho(Ay,Ax)quadforall x, yin K$$ or $$rho(x,Ax)-rho(y,Ay)lerho(x,y)+rho(Ax,Ay).$$ Reversing the roles of $x,y$ to get $$left|rho(x,Ax)-rho(y,Ay)right|lerho(x,y)+rho(Ax,Ay)<2delta quad text{ whenever }rho(x,y)<delta.$$ That is, $f$ is actually uniformly continuous.)
Let $alpha:=inf_{xin K}f(x)$, then we can find $x_0in K$ such that $alpha=f(x_0)$, since $K$ is compact. If $alpha>0$, then $x_0neq Ax_0$ and $rho(A(Ax_0),Ax_0)<rho(Ax_0,x_0)=alpha$, which is a contradiction. So $alpha=0$ and $x_0$ is a fixed point. The assumption on $A$ makes it unique.
Note that completeness wouldn't be enough in this case, for example consider $mathbb R$ with the usual metric, and $A(x):=sqrt{x^2+1}$. It's the major difference between $rho(Ax,Ay)<rho(x,y)$ for $xneq y$ and the existence of $0<c<1$ such that for all $x,y,$: $rho(Ax,Ay)leq crho(x,y)$.
edited Oct 15 '14 at 17:46
Fang Jing
7461418
answered Mar 10 '12 at 13:15
Davide Giraudo
123k16149253
edited Oct 15 '14 at 17:46
Fang Jing
7461418
edited Oct 15 '14 at 17:46
Fang Jing
7461418
edited Oct 15 '14 at 17:46
Fang Jing
7461418
7461418
answered Mar 10 '12 at 13:15
Davide Giraudo
123k16149253
answered Mar 10 '12 at 13:15
Davide Giraudo
123k16149253
answered Mar 10 '12 at 13:15
Davide Giraudo
123k16149253
123k16149253
Nice proof!Thank you! :))
– Jiangnan Yu
Mar 10 '12 at 13:53
How do we show that f(x):=ρ(x,A(x)) is indeed continuous?
– Jacques
Apr 2 '12 at 9:22
3
@Jacques: $delta: x mapsto (x,x)$ is continuous, $A$ is continuous, so $g:(x,y) mapsto (x,A(y))$ is continuous, and $d:(x,y) mapsto d(x,y)$ is continuous, so $f(x) = (dcirc g circ delta)(x)$ is a composition of continuous maps, hence it is continuous. Alternatively, use the triangle inequality and the reverse triangle inequality a few times.
– t.b.
Apr 2 '12 at 9:52
Can someone clarify about uniqueness?
– Niebla
Nov 9 '15 at 2:57
1
@Niebla In general if we have $rho(A(x), A(y))<rho(x,y)$ - note that the inequality is strict - $A$ can only have one fixed point. Let $a, b$ be two fixed points, then $rho(A(a), B(b))<rho(a, b)$, which is a contradiction since both sides of this strict inequality are equal.
– Jack M
Dec 27 '15 at 16:53
|
show 1 more comment
Nice proof!Thank you! :))
– Jiangnan Yu
Mar 10 '12 at 13:53
How do we show that f(x):=ρ(x,A(x)) is indeed continuous?
– Jacques
Apr 2 '12 at 9:22
3
@Jacques: $delta: x mapsto (x,x)$ is continuous, $A$ is continuous, so $g:(x,y) mapsto (x,A(y))$ is continuous, and $d:(x,y) mapsto d(x,y)$ is continuous, so $f(x) = (dcirc g circ delta)(x)$ is a composition of continuous maps, hence it is continuous. Alternatively, use the triangle inequality and the reverse triangle inequality a few times.
– t.b.
Apr 2 '12 at 9:52
Can someone clarify about uniqueness?
– Niebla
Nov 9 '15 at 2:57
1
@Niebla In general if we have $rho(A(x), A(y))<rho(x,y)$ - note that the inequality is strict - $A$ can only have one fixed point. Let $a, b$ be two fixed points, then $rho(A(a), B(b))<rho(a, b)$, which is a contradiction since both sides of this strict inequality are equal.
– Jack M
Dec 27 '15 at 16:53
Nice proof!Thank you! :))
– Jiangnan Yu
Mar 10 '12 at 13:53
Nice proof!Thank you! :))
– Jiangnan Yu
Mar 10 '12 at 13:53
How do we show that f(x):=ρ(x,A(x)) is indeed continuous?
– Jacques
Apr 2 '12 at 9:22
How do we show that f(x):=ρ(x,A(x)) is indeed continuous?
– Jacques
Apr 2 '12 at 9:22
3
3
@Jacques: $delta: x mapsto (x,x)$ is continuous, $A$ is continuous, so $g:(x,y) mapsto (x,A(y))$ is continuous, and $d:(x,y) mapsto d(x,y)$ is continuous, so $f(x) = (dcirc g circ delta)(x)$ is a composition of continuous maps, hence it is continuous. Alternatively, use the triangle inequality and the reverse triangle inequality a few times.
– t.b.
Apr 2 '12 at 9:52
@Jacques: $delta: x mapsto (x,x)$ is continuous, $A$ is continuous, so $g:(x,y) mapsto (x,A(y))$ is continuous, and $d:(x,y) mapsto d(x,y)$ is continuous, so $f(x) = (dcirc g circ delta)(x)$ is a composition of continuous maps, hence it is continuous. Alternatively, use the triangle inequality and the reverse triangle inequality a few times.
– t.b.
Apr 2 '12 at 9:52
Can someone clarify about uniqueness?
– Niebla
Nov 9 '15 at 2:57
Can someone clarify about uniqueness?
– Niebla
Nov 9 '15 at 2:57
1
1
@Niebla In general if we have $rho(A(x), A(y))<rho(x,y)$ - note that the inequality is strict - $A$ can only have one fixed point. Let $a, b$ be two fixed points, then $rho(A(a), B(b))<rho(a, b)$, which is a contradiction since both sides of this strict inequality are equal.
– Jack M
Dec 27 '15 at 16:53
@Niebla In general if we have $rho(A(x), A(y))<rho(x,y)$ - note that the inequality is strict - $A$ can only have one fixed point. Let $a, b$ be two fixed points, then $rho(A(a), B(b))<rho(a, b)$, which is a contradiction since both sides of this strict inequality are equal.
– Jack M
Dec 27 '15 at 16:53
|
show 1 more comment
up vote
2
down vote
I don't have enough reputation to post a comment to reply to @андрэ 's question regarding where in the proof it is used that $f$ is a continuous function, so I'll post my answer here:
Since we are told that $K$ is a compact set. $f:Krightarrow K$ being continuous implies that the $mathrm{im}(f) = f(K)$ is also a compact set. We also know that compact sets are closed and bounded, which implies the existence of $inf_{xin K} f(x)$.
If it is possible to show that $f(K) subseteq K$ is a closed set, then it is necessarily compact as well:
A subset of a compact set is compact?
However, I am not aware of how you would do this in this case without relying on continuity of $f$.
answered Oct 21 at 8:49
Matthew O'Brien
415
add a comment |
up vote
2
down vote
I don't have enough reputation to post a comment to reply to @андрэ 's question regarding where in the proof it is used that $f$ is a continuous function, so I'll post my answer here:
Since we are told that $K$ is a compact set. $f:Krightarrow K$ being continuous implies that the $mathrm{im}(f) = f(K)$ is also a compact set. We also know that compact sets are closed and bounded, which implies the existence of $inf_{xin K} f(x)$.
If it is possible to show that $f(K) subseteq K$ is a closed set, then it is necessarily compact as well:
A subset of a compact set is compact?
However, I am not aware of how you would do this in this case without relying on continuity of $f$.
answered Oct 21 at 8:49
Matthew O'Brien
415
add a comment |
up vote
2
down vote
up vote
2
down vote
I don't have enough reputation to post a comment to reply to @андрэ 's question regarding where in the proof it is used that $f$ is a continuous function, so I'll post my answer here:
Since we are told that $K$ is a compact set. $f:Krightarrow K$ being continuous implies that the $mathrm{im}(f) = f(K)$ is also a compact set. We also know that compact sets are closed and bounded, which implies the existence of $inf_{xin K} f(x)$.
If it is possible to show that $f(K) subseteq K$ is a closed set, then it is necessarily compact as well:
A subset of a compact set is compact?
However, I am not aware of how you would do this in this case without relying on continuity of $f$.
answered Oct 21 at 8:49
Matthew O'Brien
415
I don't have enough reputation to post a comment to reply to @андрэ 's question regarding where in the proof it is used that $f$ is a continuous function, so I'll post my answer here:
Since we are told that $K$ is a compact set. $f:Krightarrow K$ being continuous implies that the $mathrm{im}(f) = f(K)$ is also a compact set. We also know that compact sets are closed and bounded, which implies the existence of $inf_{xin K} f(x)$.
If it is possible to show that $f(K) subseteq K$ is a closed set, then it is necessarily compact as well:
A subset of a compact set is compact?
However, I am not aware of how you would do this in this case without relying on continuity of $f$.
answered Oct 21 at 8:49
Matthew O'Brien
415
answered Oct 21 at 8:49
Matthew O'Brien
415
answered Oct 21 at 8:49
Matthew O'Brien
415
answered Oct 21 at 8:49
Matthew O'Brien
415
415
add a comment |
add a comment |
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1
(1) I think you need to assume $K$ is complete. (2) You have a convergent subsequence; the only thing you can do now is examine the behavior of its limit ...
– Neal
Mar 10 '12 at 13:09
6
@Neal: A metric space is compact iff it is complete and totally bounded, so completeness comes for free with compactness.
– Brian M. Scott
Mar 10 '12 at 13:17
Oh, I totally missed "compact" in the question. My bad.
– Neal
Mar 11 '12 at 0:24