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Is the set of bounded functions connected?
Space of bounded continuous functions is completeAn interval is path connectedProving that $S^n$ (n-sphere) is locally connected.$X$ is locally connected and $f: X to Y$ is onto where $Y$ has the quotient topology. Prove that $Y$ is locally connectedWhy is the trivial topology on $X$ simply connected?Is a Normed Vector Space Necessary to Prove Path Connectedness?Connected components and continuous functionsHelp with a proof; Path connected open subspaces imply $X$ to be path connectedIs Every (Non-Trivial) Path Connected Space Uncountable?Is a set with the cofinite topology path-wise connected?$G_f = (x,y) in X times Y vert y = f(x) $ connected $Leftrightarrow X$ connected
$begingroup$
One of my topology homework questions this week says the following:
Consider the space of bounded functions $B[0, 1]$ on the interval $[0, 1]$
with the $dinfty$ metric.
Prove that in the topology induced by the metric $B[0, 1]$ is connected.
I am not sure how to go about proving this, I tried showing that it is path-connected, but since the functions are not necessarily continuous I can't construct a continuous path between them.
The closest answer I've found so far can be found here, but I'm not sure how to apply it to my problem, again since the functions I'm dealing with are not continuous.
general-topology metric-spaces connectedness
edited Mar 15 at 16:56
Max
9051318
asked Mar 15 at 16:36
niknik
184
$endgroup$
add a comment |
$begingroup$
One of my topology homework questions this week says the following:
Consider the space of bounded functions $B[0, 1]$ on the interval $[0, 1]$
with the $dinfty$ metric.
Prove that in the topology induced by the metric $B[0, 1]$ is connected.
I am not sure how to go about proving this, I tried showing that it is path-connected, but since the functions are not necessarily continuous I can't construct a continuous path between them.
The closest answer I've found so far can be found here, but I'm not sure how to apply it to my problem, again since the functions I'm dealing with are not continuous.
general-topology metric-spaces connectedness
edited Mar 15 at 16:56
Max
9051318
asked Mar 15 at 16:36
niknik
184
$endgroup$
1
$begingroup$
You don't need to construct a path of continuous between them. You just need to construct a path of bounded functions between them. My guess is that whatever formula you might wish to use for a continuous path of continuous functions would work perfectly for a continuous path of bounded functions.
$endgroup$
– Lee Mosher
Mar 15 at 16:58
add a comment |
$begingroup$
One of my topology homework questions this week says the following:
Consider the space of bounded functions $B[0, 1]$ on the interval $[0, 1]$
with the $dinfty$ metric.
Prove that in the topology induced by the metric $B[0, 1]$ is connected.
I am not sure how to go about proving this, I tried showing that it is path-connected, but since the functions are not necessarily continuous I can't construct a continuous path between them.
The closest answer I've found so far can be found here, but I'm not sure how to apply it to my problem, again since the functions I'm dealing with are not continuous.
general-topology metric-spaces connectedness
edited Mar 15 at 16:56
Max
9051318
asked Mar 15 at 16:36
niknik
184
$endgroup$
One of my topology homework questions this week says the following:
Consider the space of bounded functions $B[0, 1]$ on the interval $[0, 1]$
with the $dinfty$ metric.
Prove that in the topology induced by the metric $B[0, 1]$ is connected.
I am not sure how to go about proving this, I tried showing that it is path-connected, but since the functions are not necessarily continuous I can't construct a continuous path between them.
The closest answer I've found so far can be found here, but I'm not sure how to apply it to my problem, again since the functions I'm dealing with are not continuous.
general-topology metric-spaces connectedness
general-topology metric-spaces connectedness
edited Mar 15 at 16:56
Max
9051318
asked Mar 15 at 16:36
niknik
184
edited Mar 15 at 16:56
Max
9051318
asked Mar 15 at 16:36
niknik
184
edited Mar 15 at 16:56
Max
9051318
edited Mar 15 at 16:56
Max
9051318
edited Mar 15 at 16:56
Max
9051318
9051318
asked Mar 15 at 16:36
niknik
184
asked Mar 15 at 16:36
niknik
184
asked Mar 15 at 16:36
niknik
184
184
1
$begingroup$
You don't need to construct a path of continuous between them. You just need to construct a path of bounded functions between them. My guess is that whatever formula you might wish to use for a continuous path of continuous functions would work perfectly for a continuous path of bounded functions.
$endgroup$
– Lee Mosher
Mar 15 at 16:58
add a comment |
1
$begingroup$
You don't need to construct a path of continuous between them. You just need to construct a path of bounded functions between them. My guess is that whatever formula you might wish to use for a continuous path of continuous functions would work perfectly for a continuous path of bounded functions.
$endgroup$
– Lee Mosher
Mar 15 at 16:58
1
1
$begingroup$
You don't need to construct a path of continuous between them. You just need to construct a path of bounded functions between them. My guess is that whatever formula you might wish to use for a continuous path of continuous functions would work perfectly for a continuous path of bounded functions.
$endgroup$
– Lee Mosher
Mar 15 at 16:58
$begingroup$
You don't need to construct a path of continuous between them. You just need to construct a path of bounded functions between them. My guess is that whatever formula you might wish to use for a continuous path of continuous functions would work perfectly for a continuous path of bounded functions.
$endgroup$
– Lee Mosher
Mar 15 at 16:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $f, g in B([0,1])$. By definition, there exists $M, M'$ such that for all $x in [0,1]$,
$$|f(x)| leq M quad textand quad |g(x)| leq M'$$
For all $t in [0,1]$, consider the function $f_t : [0,1] rightarrow mathbbR$ defined by
$$f_t : x mapsto tf(x) + (1-t)g(x)$$
First, $f_t in B([0,1])$ for all $t in [0,1]$. Indeed it is easy to see that $f_t$ is bounded by $max(M,M')$.
Moreover, $f_0 = g$ and $f_1 = f$. To see that $t mapsto f_t$ defines a path between $g$ and $f$, you have to prove that it is continuous. Let $varepsilon > 0$. Define $$eta = fracvarepsilonM+M' +1$$
If $|t'-t| leq eta$, then
$$d_infty(f_t,f_t') = sup_x in [0,1] |f_t(x) - f_t'(x)| = sup_x in [0,1] | (t-t')f(x) + (t'-t)g(x)|$$ $$ leq |t-t'| (M+M') leq fracvarepsilonM+M' +1 (M+M') leq varepsilon$$
This shows that $t mapsto f_t$ is continuous, and therefore, $B([0,1])$ is path-connected.
answered Mar 15 at 16:48
TheSilverDoeTheSilverDoe
4,523114
$endgroup$
add a comment |
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$begingroup$
Let $f, g in B([0,1])$. By definition, there exists $M, M'$ such that for all $x in [0,1]$,
$$|f(x)| leq M quad textand quad |g(x)| leq M'$$
For all $t in [0,1]$, consider the function $f_t : [0,1] rightarrow mathbbR$ defined by
$$f_t : x mapsto tf(x) + (1-t)g(x)$$
First, $f_t in B([0,1])$ for all $t in [0,1]$. Indeed it is easy to see that $f_t$ is bounded by $max(M,M')$.
Moreover, $f_0 = g$ and $f_1 = f$. To see that $t mapsto f_t$ defines a path between $g$ and $f$, you have to prove that it is continuous. Let $varepsilon > 0$. Define $$eta = fracvarepsilonM+M' +1$$
If $|t'-t| leq eta$, then
$$d_infty(f_t,f_t') = sup_x in [0,1] |f_t(x) - f_t'(x)| = sup_x in [0,1] | (t-t')f(x) + (t'-t)g(x)|$$ $$ leq |t-t'| (M+M') leq fracvarepsilonM+M' +1 (M+M') leq varepsilon$$
This shows that $t mapsto f_t$ is continuous, and therefore, $B([0,1])$ is path-connected.
answered Mar 15 at 16:48
TheSilverDoeTheSilverDoe
4,523114
$endgroup$
add a comment |
$begingroup$
Let $f, g in B([0,1])$. By definition, there exists $M, M'$ such that for all $x in [0,1]$,
$$|f(x)| leq M quad textand quad |g(x)| leq M'$$
For all $t in [0,1]$, consider the function $f_t : [0,1] rightarrow mathbbR$ defined by
$$f_t : x mapsto tf(x) + (1-t)g(x)$$
First, $f_t in B([0,1])$ for all $t in [0,1]$. Indeed it is easy to see that $f_t$ is bounded by $max(M,M')$.
Moreover, $f_0 = g$ and $f_1 = f$. To see that $t mapsto f_t$ defines a path between $g$ and $f$, you have to prove that it is continuous. Let $varepsilon > 0$. Define $$eta = fracvarepsilonM+M' +1$$
If $|t'-t| leq eta$, then
$$d_infty(f_t,f_t') = sup_x in [0,1] |f_t(x) - f_t'(x)| = sup_x in [0,1] | (t-t')f(x) + (t'-t)g(x)|$$ $$ leq |t-t'| (M+M') leq fracvarepsilonM+M' +1 (M+M') leq varepsilon$$
This shows that $t mapsto f_t$ is continuous, and therefore, $B([0,1])$ is path-connected.
answered Mar 15 at 16:48
TheSilverDoeTheSilverDoe
4,523114
$endgroup$
add a comment |
$begingroup$
Let $f, g in B([0,1])$. By definition, there exists $M, M'$ such that for all $x in [0,1]$,
$$|f(x)| leq M quad textand quad |g(x)| leq M'$$
For all $t in [0,1]$, consider the function $f_t : [0,1] rightarrow mathbbR$ defined by
$$f_t : x mapsto tf(x) + (1-t)g(x)$$
First, $f_t in B([0,1])$ for all $t in [0,1]$. Indeed it is easy to see that $f_t$ is bounded by $max(M,M')$.
Moreover, $f_0 = g$ and $f_1 = f$. To see that $t mapsto f_t$ defines a path between $g$ and $f$, you have to prove that it is continuous. Let $varepsilon > 0$. Define $$eta = fracvarepsilonM+M' +1$$
If $|t'-t| leq eta$, then
$$d_infty(f_t,f_t') = sup_x in [0,1] |f_t(x) - f_t'(x)| = sup_x in [0,1] | (t-t')f(x) + (t'-t)g(x)|$$ $$ leq |t-t'| (M+M') leq fracvarepsilonM+M' +1 (M+M') leq varepsilon$$
This shows that $t mapsto f_t$ is continuous, and therefore, $B([0,1])$ is path-connected.
answered Mar 15 at 16:48
TheSilverDoeTheSilverDoe
4,523114
$endgroup$
Let $f, g in B([0,1])$. By definition, there exists $M, M'$ such that for all $x in [0,1]$,
$$|f(x)| leq M quad textand quad |g(x)| leq M'$$
For all $t in [0,1]$, consider the function $f_t : [0,1] rightarrow mathbbR$ defined by
$$f_t : x mapsto tf(x) + (1-t)g(x)$$
First, $f_t in B([0,1])$ for all $t in [0,1]$. Indeed it is easy to see that $f_t$ is bounded by $max(M,M')$.
Moreover, $f_0 = g$ and $f_1 = f$. To see that $t mapsto f_t$ defines a path between $g$ and $f$, you have to prove that it is continuous. Let $varepsilon > 0$. Define $$eta = fracvarepsilonM+M' +1$$
If $|t'-t| leq eta$, then
$$d_infty(f_t,f_t') = sup_x in [0,1] |f_t(x) - f_t'(x)| = sup_x in [0,1] | (t-t')f(x) + (t'-t)g(x)|$$ $$ leq |t-t'| (M+M') leq fracvarepsilonM+M' +1 (M+M') leq varepsilon$$
This shows that $t mapsto f_t$ is continuous, and therefore, $B([0,1])$ is path-connected.
answered Mar 15 at 16:48
TheSilverDoeTheSilverDoe
4,523114
answered Mar 15 at 16:48
TheSilverDoeTheSilverDoe
4,523114
answered Mar 15 at 16:48
TheSilverDoeTheSilverDoe
4,523114
answered Mar 15 at 16:48
TheSilverDoeTheSilverDoe
4,523114
4,523114
add a comment |
add a comment |
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$begingroup$
You don't need to construct a path of continuous between them. You just need to construct a path of bounded functions between them. My guess is that whatever formula you might wish to use for a continuous path of continuous functions would work perfectly for a continuous path of bounded functions.
$endgroup$
– Lee Mosher
Mar 15 at 16:58