Mixing
$f$ is continuous on $E$ if and only if its graph is compact.
$begingroup$
This question may be asked before under different formulation, the original problem is Chapter 4, Exercise 7 of Rudin's text: The Principles of Mathematical Analysis:
Problem:
If $f$ is defined on $E$, the graph is the set of points $(x, f(x))$, for $x in E$.
Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.
My Attempt: Let $Gamma(f) = {(x, f(x)): x in E} subset E times mathbb{R}^1$. I was trying to define a function $F: E rightarrow E times mathbb{R}^1$ by $F(x) = (x, f(x))$ and show $F$ is continuous on $E$. Since $E$ is compact, it follows that $F(E) = Gamma(f)$ is compact.
For the converse, the earlier thread
A real function on a compact set is continuous if and only if its graph is compact stated that the projection function $pi$ is continuous on $E times mathbb{R}^1$.
My question is: without knowing any specific metric on $E$ or on $E times mathbb{R}^1$, it looks hard to me to show that $F$ and $pi$ aforementioned are continuous. Since the condition doesn't give any information about what I concern, how should I proceed?
general-topology analysis
edited Apr 13 '17 at 12:21
Community♦
1
asked Jul 6 '15 at 22:39
ZhanxiongZhanxiong
8,88111031
$endgroup$
|
show 4 more comments
$begingroup$
This question may be asked before under different formulation, the original problem is Chapter 4, Exercise 7 of Rudin's text: The Principles of Mathematical Analysis:
Problem:
If $f$ is defined on $E$, the graph is the set of points $(x, f(x))$, for $x in E$.
Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.
My Attempt: Let $Gamma(f) = {(x, f(x)): x in E} subset E times mathbb{R}^1$. I was trying to define a function $F: E rightarrow E times mathbb{R}^1$ by $F(x) = (x, f(x))$ and show $F$ is continuous on $E$. Since $E$ is compact, it follows that $F(E) = Gamma(f)$ is compact.
For the converse, the earlier thread
A real function on a compact set is continuous if and only if its graph is compact stated that the projection function $pi$ is continuous on $E times mathbb{R}^1$.
My question is: without knowing any specific metric on $E$ or on $E times mathbb{R}^1$, it looks hard to me to show that $F$ and $pi$ aforementioned are continuous. Since the condition doesn't give any information about what I concern, how should I proceed?
general-topology analysis
edited Apr 13 '17 at 12:21
Community♦
1
asked Jul 6 '15 at 22:39
ZhanxiongZhanxiong
8,88111031
$endgroup$
1
$begingroup$
We can get a metric on this product by defining an induced metric on the product of metric spaces. For example, given $(X, d_X)$ and $(Y,d_Y)$, it is common to define the induced metric on $X times Y$ by $$ d_{X times Y}((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2) $$
$endgroup$
– Omnomnomnom
Jul 6 '15 at 22:47
$begingroup$
@Omnomnomnom Thank you for your reply. I knew we might can do so, but I am actually trying to avoid such "subjective" proof. I am wondering is there any "objective" proof that without imposing any metric by ourselves? I think the existence of $d_X$ is fine, the annoying part is can we proceed without giving any specific metric form on $d_{X times mathbb{R}^1}$?
$endgroup$
– Zhanxiong
Jul 6 '15 at 23:02
$begingroup$
the problem is that you have to say what it means for $Gamma(f)$ to be compact, for which we need some kind of topology on $E times Bbb R$. If you're not going to get this topology using a "subjective" choice of metric on the product, then perhaps you should do so using the definition of the product topology, ignoring the underlying metrics altogether.
$endgroup$
– Omnomnomnom
Jul 6 '15 at 23:06
$begingroup$
@Zhanxiong the metric .@Omnomnomnom defined induces the product topology, as there are only finitely many factors. So it isn't "subjective". Whatever that in this context might mean.
$endgroup$
– user251257
Jul 6 '15 at 23:13
$begingroup$
So literally, we can use any metric on $E times mathbb{R}^1$, such as $d_{E times mathbb{R}^1}((x_1, y_1), (x_2, y_2)) = max(d_E(x_1, x_2), |y_1 - y_2|)$, right? What I mean by "subjective" is like this --- the proof details depend on the specification of the metric on $E times mathbb{R}^1$, which I understand is acceptable but not that perfect...
$endgroup$
– Zhanxiong
Jul 7 '15 at 1:20
|
show 4 more comments
$begingroup$
This question may be asked before under different formulation, the original problem is Chapter 4, Exercise 7 of Rudin's text: The Principles of Mathematical Analysis:
Problem:
If $f$ is defined on $E$, the graph is the set of points $(x, f(x))$, for $x in E$.
Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.
My Attempt: Let $Gamma(f) = {(x, f(x)): x in E} subset E times mathbb{R}^1$. I was trying to define a function $F: E rightarrow E times mathbb{R}^1$ by $F(x) = (x, f(x))$ and show $F$ is continuous on $E$. Since $E$ is compact, it follows that $F(E) = Gamma(f)$ is compact.
For the converse, the earlier thread
A real function on a compact set is continuous if and only if its graph is compact stated that the projection function $pi$ is continuous on $E times mathbb{R}^1$.
My question is: without knowing any specific metric on $E$ or on $E times mathbb{R}^1$, it looks hard to me to show that $F$ and $pi$ aforementioned are continuous. Since the condition doesn't give any information about what I concern, how should I proceed?
general-topology analysis
edited Apr 13 '17 at 12:21
Community♦
1
asked Jul 6 '15 at 22:39
ZhanxiongZhanxiong
8,88111031
$endgroup$
This question may be asked before under different formulation, the original problem is Chapter 4, Exercise 7 of Rudin's text: The Principles of Mathematical Analysis:
Problem:
If $f$ is defined on $E$, the graph is the set of points $(x, f(x))$, for $x in E$.
Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.
My Attempt: Let $Gamma(f) = {(x, f(x)): x in E} subset E times mathbb{R}^1$. I was trying to define a function $F: E rightarrow E times mathbb{R}^1$ by $F(x) = (x, f(x))$ and show $F$ is continuous on $E$. Since $E$ is compact, it follows that $F(E) = Gamma(f)$ is compact.
For the converse, the earlier thread
A real function on a compact set is continuous if and only if its graph is compact stated that the projection function $pi$ is continuous on $E times mathbb{R}^1$.
My question is: without knowing any specific metric on $E$ or on $E times mathbb{R}^1$, it looks hard to me to show that $F$ and $pi$ aforementioned are continuous. Since the condition doesn't give any information about what I concern, how should I proceed?
general-topology analysis
general-topology analysis
edited Apr 13 '17 at 12:21
Community♦
1
asked Jul 6 '15 at 22:39
ZhanxiongZhanxiong
8,88111031
edited Apr 13 '17 at 12:21
Community♦
1
asked Jul 6 '15 at 22:39
ZhanxiongZhanxiong
8,88111031
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Jul 6 '15 at 22:39
ZhanxiongZhanxiong
8,88111031
asked Jul 6 '15 at 22:39
ZhanxiongZhanxiong
8,88111031
asked Jul 6 '15 at 22:39
ZhanxiongZhanxiong
8,88111031
8,88111031
1
$begingroup$
We can get a metric on this product by defining an induced metric on the product of metric spaces. For example, given $(X, d_X)$ and $(Y,d_Y)$, it is common to define the induced metric on $X times Y$ by $$ d_{X times Y}((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2) $$
$endgroup$
– Omnomnomnom
Jul 6 '15 at 22:47
$begingroup$
@Omnomnomnom Thank you for your reply. I knew we might can do so, but I am actually trying to avoid such "subjective" proof. I am wondering is there any "objective" proof that without imposing any metric by ourselves? I think the existence of $d_X$ is fine, the annoying part is can we proceed without giving any specific metric form on $d_{X times mathbb{R}^1}$?
$endgroup$
– Zhanxiong
Jul 6 '15 at 23:02
$begingroup$
the problem is that you have to say what it means for $Gamma(f)$ to be compact, for which we need some kind of topology on $E times Bbb R$. If you're not going to get this topology using a "subjective" choice of metric on the product, then perhaps you should do so using the definition of the product topology, ignoring the underlying metrics altogether.
$endgroup$
– Omnomnomnom
Jul 6 '15 at 23:06
$begingroup$
@Zhanxiong the metric .@Omnomnomnom defined induces the product topology, as there are only finitely many factors. So it isn't "subjective". Whatever that in this context might mean.
$endgroup$
– user251257
Jul 6 '15 at 23:13
$begingroup$
So literally, we can use any metric on $E times mathbb{R}^1$, such as $d_{E times mathbb{R}^1}((x_1, y_1), (x_2, y_2)) = max(d_E(x_1, x_2), |y_1 - y_2|)$, right? What I mean by "subjective" is like this --- the proof details depend on the specification of the metric on $E times mathbb{R}^1$, which I understand is acceptable but not that perfect...
$endgroup$
– Zhanxiong
Jul 7 '15 at 1:20
|
show 4 more comments
1
$begingroup$
We can get a metric on this product by defining an induced metric on the product of metric spaces. For example, given $(X, d_X)$ and $(Y,d_Y)$, it is common to define the induced metric on $X times Y$ by $$ d_{X times Y}((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2) $$
$endgroup$
– Omnomnomnom
Jul 6 '15 at 22:47
$begingroup$
@Omnomnomnom Thank you for your reply. I knew we might can do so, but I am actually trying to avoid such "subjective" proof. I am wondering is there any "objective" proof that without imposing any metric by ourselves? I think the existence of $d_X$ is fine, the annoying part is can we proceed without giving any specific metric form on $d_{X times mathbb{R}^1}$?
$endgroup$
– Zhanxiong
Jul 6 '15 at 23:02
$begingroup$
the problem is that you have to say what it means for $Gamma(f)$ to be compact, for which we need some kind of topology on $E times Bbb R$. If you're not going to get this topology using a "subjective" choice of metric on the product, then perhaps you should do so using the definition of the product topology, ignoring the underlying metrics altogether.
$endgroup$
– Omnomnomnom
Jul 6 '15 at 23:06
$begingroup$
@Zhanxiong the metric .@Omnomnomnom defined induces the product topology, as there are only finitely many factors. So it isn't "subjective". Whatever that in this context might mean.
$endgroup$
– user251257
Jul 6 '15 at 23:13
$begingroup$
So literally, we can use any metric on $E times mathbb{R}^1$, such as $d_{E times mathbb{R}^1}((x_1, y_1), (x_2, y_2)) = max(d_E(x_1, x_2), |y_1 - y_2|)$, right? What I mean by "subjective" is like this --- the proof details depend on the specification of the metric on $E times mathbb{R}^1$, which I understand is acceptable but not that perfect...
$endgroup$
– Zhanxiong
Jul 7 '15 at 1:20
1
1
$begingroup$
We can get a metric on this product by defining an induced metric on the product of metric spaces. For example, given $(X, d_X)$ and $(Y,d_Y)$, it is common to define the induced metric on $X times Y$ by $$ d_{X times Y}((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2) $$
$endgroup$
– Omnomnomnom
Jul 6 '15 at 22:47
$begingroup$
We can get a metric on this product by defining an induced metric on the product of metric spaces. For example, given $(X, d_X)$ and $(Y,d_Y)$, it is common to define the induced metric on $X times Y$ by $$ d_{X times Y}((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2) $$
$endgroup$
– Omnomnomnom
Jul 6 '15 at 22:47
$begingroup$
@Omnomnomnom Thank you for your reply. I knew we might can do so, but I am actually trying to avoid such "subjective" proof. I am wondering is there any "objective" proof that without imposing any metric by ourselves? I think the existence of $d_X$ is fine, the annoying part is can we proceed without giving any specific metric form on $d_{X times mathbb{R}^1}$?
$endgroup$
– Zhanxiong
Jul 6 '15 at 23:02
$begingroup$
@Omnomnomnom Thank you for your reply. I knew we might can do so, but I am actually trying to avoid such "subjective" proof. I am wondering is there any "objective" proof that without imposing any metric by ourselves? I think the existence of $d_X$ is fine, the annoying part is can we proceed without giving any specific metric form on $d_{X times mathbb{R}^1}$?
$endgroup$
– Zhanxiong
Jul 6 '15 at 23:02
$begingroup$
the problem is that you have to say what it means for $Gamma(f)$ to be compact, for which we need some kind of topology on $E times Bbb R$. If you're not going to get this topology using a "subjective" choice of metric on the product, then perhaps you should do so using the definition of the product topology, ignoring the underlying metrics altogether.
$endgroup$
– Omnomnomnom
Jul 6 '15 at 23:06
$begingroup$
the problem is that you have to say what it means for $Gamma(f)$ to be compact, for which we need some kind of topology on $E times Bbb R$. If you're not going to get this topology using a "subjective" choice of metric on the product, then perhaps you should do so using the definition of the product topology, ignoring the underlying metrics altogether.
$endgroup$
– Omnomnomnom
Jul 6 '15 at 23:06
$begingroup$
@Zhanxiong the metric .@Omnomnomnom defined induces the product topology, as there are only finitely many factors. So it isn't "subjective". Whatever that in this context might mean.
$endgroup$
– user251257
Jul 6 '15 at 23:13
$begingroup$
@Zhanxiong the metric .@Omnomnomnom defined induces the product topology, as there are only finitely many factors. So it isn't "subjective". Whatever that in this context might mean.
$endgroup$
– user251257
Jul 6 '15 at 23:13
$begingroup$
So literally, we can use any metric on $E times mathbb{R}^1$, such as $d_{E times mathbb{R}^1}((x_1, y_1), (x_2, y_2)) = max(d_E(x_1, x_2), |y_1 - y_2|)$, right? What I mean by "subjective" is like this --- the proof details depend on the specification of the metric on $E times mathbb{R}^1$, which I understand is acceptable but not that perfect...
$endgroup$
– Zhanxiong
Jul 7 '15 at 1:20
$begingroup$
So literally, we can use any metric on $E times mathbb{R}^1$, such as $d_{E times mathbb{R}^1}((x_1, y_1), (x_2, y_2)) = max(d_E(x_1, x_2), |y_1 - y_2|)$, right? What I mean by "subjective" is like this --- the proof details depend on the specification of the metric on $E times mathbb{R}^1$, which I understand is acceptable but not that perfect...
$endgroup$
– Zhanxiong
Jul 7 '15 at 1:20
|
show 4 more comments
1 Answer
1
active
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$begingroup$
Suppose that $f$ is continuous, and define $F(x):=(x,f(x))subset Etimesmathbb R$. Fix $xin E$ and consider a sequence $(x_n)subset E$ converging to $x$. Since $f$ is continuous, $f(x_n)to f(x)$ and hence $F(x_n)=(x_n,f(x_n))to(x,f(x))=F(x)$, which shows that $F$ is continuous. Now your argument works fine, since now $Gamma(f)=F(E)$ is compact.
For the converse assume that $Gamma(f)=F(E)$ is compact. Fix $xin E$ and a sequence $(x_n)subset E$ converging to $x$. Assume that $(x_n,f(x_n))$ does not converge to $(x,f(x))$. Since $Gamma(f)$ is compact, there is a subsequence $(x_{n_k},f(x_{n_k})$ that converges in $Gamma(f)$ to some $(x',y)neq(x,f(x))$. But since $(x_n)$ converges to $x$, then so does $(x_{n_k})$ (because it's a subsequence), which implies $x'=x$. But since $(x',y)inGamma(f)$, we must have $y=f(x')=f(x)$ and consequently $(x',y)=(x,f(x))$, a contradiction. This shows that $(x_n,f(x_n))to(x,f(x))$; in particular, $f(x_n)to f(x)$, which proves that $f$ is continuous at $x$ (and that $F$ is continuous at $x$ as well).
answered Jul 6 '15 at 23:15
sranthropsranthrop
7,0841925
$endgroup$
$begingroup$
Does this generalize ? If $E$ is any compact space and $D$ is any space, is it true that $f:Eto D$ is continuous iff the graph of $f$ is compact?......+1
$endgroup$
– DanielWainfleet
Jan 9 at 20:55
add a comment |
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$begingroup$
Suppose that $f$ is continuous, and define $F(x):=(x,f(x))subset Etimesmathbb R$. Fix $xin E$ and consider a sequence $(x_n)subset E$ converging to $x$. Since $f$ is continuous, $f(x_n)to f(x)$ and hence $F(x_n)=(x_n,f(x_n))to(x,f(x))=F(x)$, which shows that $F$ is continuous. Now your argument works fine, since now $Gamma(f)=F(E)$ is compact.
For the converse assume that $Gamma(f)=F(E)$ is compact. Fix $xin E$ and a sequence $(x_n)subset E$ converging to $x$. Assume that $(x_n,f(x_n))$ does not converge to $(x,f(x))$. Since $Gamma(f)$ is compact, there is a subsequence $(x_{n_k},f(x_{n_k})$ that converges in $Gamma(f)$ to some $(x',y)neq(x,f(x))$. But since $(x_n)$ converges to $x$, then so does $(x_{n_k})$ (because it's a subsequence), which implies $x'=x$. But since $(x',y)inGamma(f)$, we must have $y=f(x')=f(x)$ and consequently $(x',y)=(x,f(x))$, a contradiction. This shows that $(x_n,f(x_n))to(x,f(x))$; in particular, $f(x_n)to f(x)$, which proves that $f$ is continuous at $x$ (and that $F$ is continuous at $x$ as well).
answered Jul 6 '15 at 23:15
sranthropsranthrop
7,0841925
$endgroup$
$begingroup$
Does this generalize ? If $E$ is any compact space and $D$ is any space, is it true that $f:Eto D$ is continuous iff the graph of $f$ is compact?......+1
$endgroup$
– DanielWainfleet
Jan 9 at 20:55
add a comment |
$begingroup$
Suppose that $f$ is continuous, and define $F(x):=(x,f(x))subset Etimesmathbb R$. Fix $xin E$ and consider a sequence $(x_n)subset E$ converging to $x$. Since $f$ is continuous, $f(x_n)to f(x)$ and hence $F(x_n)=(x_n,f(x_n))to(x,f(x))=F(x)$, which shows that $F$ is continuous. Now your argument works fine, since now $Gamma(f)=F(E)$ is compact.
For the converse assume that $Gamma(f)=F(E)$ is compact. Fix $xin E$ and a sequence $(x_n)subset E$ converging to $x$. Assume that $(x_n,f(x_n))$ does not converge to $(x,f(x))$. Since $Gamma(f)$ is compact, there is a subsequence $(x_{n_k},f(x_{n_k})$ that converges in $Gamma(f)$ to some $(x',y)neq(x,f(x))$. But since $(x_n)$ converges to $x$, then so does $(x_{n_k})$ (because it's a subsequence), which implies $x'=x$. But since $(x',y)inGamma(f)$, we must have $y=f(x')=f(x)$ and consequently $(x',y)=(x,f(x))$, a contradiction. This shows that $(x_n,f(x_n))to(x,f(x))$; in particular, $f(x_n)to f(x)$, which proves that $f$ is continuous at $x$ (and that $F$ is continuous at $x$ as well).
answered Jul 6 '15 at 23:15
sranthropsranthrop
7,0841925
$endgroup$
$begingroup$
Does this generalize ? If $E$ is any compact space and $D$ is any space, is it true that $f:Eto D$ is continuous iff the graph of $f$ is compact?......+1
$endgroup$
– DanielWainfleet
Jan 9 at 20:55
add a comment |
$begingroup$
Suppose that $f$ is continuous, and define $F(x):=(x,f(x))subset Etimesmathbb R$. Fix $xin E$ and consider a sequence $(x_n)subset E$ converging to $x$. Since $f$ is continuous, $f(x_n)to f(x)$ and hence $F(x_n)=(x_n,f(x_n))to(x,f(x))=F(x)$, which shows that $F$ is continuous. Now your argument works fine, since now $Gamma(f)=F(E)$ is compact.
For the converse assume that $Gamma(f)=F(E)$ is compact. Fix $xin E$ and a sequence $(x_n)subset E$ converging to $x$. Assume that $(x_n,f(x_n))$ does not converge to $(x,f(x))$. Since $Gamma(f)$ is compact, there is a subsequence $(x_{n_k},f(x_{n_k})$ that converges in $Gamma(f)$ to some $(x',y)neq(x,f(x))$. But since $(x_n)$ converges to $x$, then so does $(x_{n_k})$ (because it's a subsequence), which implies $x'=x$. But since $(x',y)inGamma(f)$, we must have $y=f(x')=f(x)$ and consequently $(x',y)=(x,f(x))$, a contradiction. This shows that $(x_n,f(x_n))to(x,f(x))$; in particular, $f(x_n)to f(x)$, which proves that $f$ is continuous at $x$ (and that $F$ is continuous at $x$ as well).
answered Jul 6 '15 at 23:15
sranthropsranthrop
7,0841925
$endgroup$
Suppose that $f$ is continuous, and define $F(x):=(x,f(x))subset Etimesmathbb R$. Fix $xin E$ and consider a sequence $(x_n)subset E$ converging to $x$. Since $f$ is continuous, $f(x_n)to f(x)$ and hence $F(x_n)=(x_n,f(x_n))to(x,f(x))=F(x)$, which shows that $F$ is continuous. Now your argument works fine, since now $Gamma(f)=F(E)$ is compact.
For the converse assume that $Gamma(f)=F(E)$ is compact. Fix $xin E$ and a sequence $(x_n)subset E$ converging to $x$. Assume that $(x_n,f(x_n))$ does not converge to $(x,f(x))$. Since $Gamma(f)$ is compact, there is a subsequence $(x_{n_k},f(x_{n_k})$ that converges in $Gamma(f)$ to some $(x',y)neq(x,f(x))$. But since $(x_n)$ converges to $x$, then so does $(x_{n_k})$ (because it's a subsequence), which implies $x'=x$. But since $(x',y)inGamma(f)$, we must have $y=f(x')=f(x)$ and consequently $(x',y)=(x,f(x))$, a contradiction. This shows that $(x_n,f(x_n))to(x,f(x))$; in particular, $f(x_n)to f(x)$, which proves that $f$ is continuous at $x$ (and that $F$ is continuous at $x$ as well).
answered Jul 6 '15 at 23:15
sranthropsranthrop
7,0841925
answered Jul 6 '15 at 23:15
sranthropsranthrop
7,0841925
answered Jul 6 '15 at 23:15
sranthropsranthrop
7,0841925
answered Jul 6 '15 at 23:15
sranthropsranthrop
7,0841925
7,0841925
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Does this generalize ? If $E$ is any compact space and $D$ is any space, is it true that $f:Eto D$ is continuous iff the graph of $f$ is compact?......+1
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– DanielWainfleet
Jan 9 at 20:55
add a comment |
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Does this generalize ? If $E$ is any compact space and $D$ is any space, is it true that $f:Eto D$ is continuous iff the graph of $f$ is compact?......+1
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– DanielWainfleet
Jan 9 at 20:55
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Does this generalize ? If $E$ is any compact space and $D$ is any space, is it true that $f:Eto D$ is continuous iff the graph of $f$ is compact?......+1
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– DanielWainfleet
Jan 9 at 20:55
$begingroup$
Does this generalize ? If $E$ is any compact space and $D$ is any space, is it true that $f:Eto D$ is continuous iff the graph of $f$ is compact?......+1
$endgroup$
– DanielWainfleet
Jan 9 at 20:55
add a comment |
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1
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We can get a metric on this product by defining an induced metric on the product of metric spaces. For example, given $(X, d_X)$ and $(Y,d_Y)$, it is common to define the induced metric on $X times Y$ by $$ d_{X times Y}((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2) $$
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– Omnomnomnom
Jul 6 '15 at 22:47
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@Omnomnomnom Thank you for your reply. I knew we might can do so, but I am actually trying to avoid such "subjective" proof. I am wondering is there any "objective" proof that without imposing any metric by ourselves? I think the existence of $d_X$ is fine, the annoying part is can we proceed without giving any specific metric form on $d_{X times mathbb{R}^1}$?
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– Zhanxiong
Jul 6 '15 at 23:02
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the problem is that you have to say what it means for $Gamma(f)$ to be compact, for which we need some kind of topology on $E times Bbb R$. If you're not going to get this topology using a "subjective" choice of metric on the product, then perhaps you should do so using the definition of the product topology, ignoring the underlying metrics altogether.
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– Omnomnomnom
Jul 6 '15 at 23:06
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@Zhanxiong the metric .@Omnomnomnom defined induces the product topology, as there are only finitely many factors. So it isn't "subjective". Whatever that in this context might mean.
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– user251257
Jul 6 '15 at 23:13
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So literally, we can use any metric on $E times mathbb{R}^1$, such as $d_{E times mathbb{R}^1}((x_1, y_1), (x_2, y_2)) = max(d_E(x_1, x_2), |y_1 - y_2|)$, right? What I mean by "subjective" is like this --- the proof details depend on the specification of the metric on $E times mathbb{R}^1$, which I understand is acceptable but not that perfect...
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– Zhanxiong
Jul 7 '15 at 1:20