Is a function on a ball of radius 1 around the zero vector bounded if it is continuous?Continuous linear mapping and bounded subsetsHow can we say that closure of an open ball equals to closed ball with the same radius on the Euclidean space?Definition of bounded set in a topological vector spaceWhen is a metrizable topological vector space locally bounded?Can the ball $B(0,r_0)$ be covered with a finite number of balls of radius $<r_0$Proving set of bounded continuous functions is an open setGeometrical representation of the unit ball?Open ball around a setIs the ball compact?open ball definition and bounded
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Is a function on a ball of radius 1 around the zero vector bounded if it is continuous?
Continuous linear mapping and bounded subsetsHow can we say that closure of an open ball equals to closed ball with the same radius on the Euclidean space?Definition of bounded set in a topological vector spaceWhen is a metrizable topological vector space locally bounded?Can the ball $B(0,r_0)$ be covered with a finite number of balls of radius $<r_0$Proving set of bounded continuous functions is an open setGeometrical representation of the unit ball?Open ball around a setIs the ball compact?open ball definition and bounded
$begingroup$
Is a function on a ball of radius 1 around the zero vector bounded if it is continuous?
If yes/no, why is that?
general-topology functional-analysis
asked Mar 10 at 19:48
2000mroliver2000mroliver
644
$endgroup$
|
show 1 more comment
$begingroup$
Is a function on a ball of radius 1 around the zero vector bounded if it is continuous?
If yes/no, why is that?
general-topology functional-analysis
asked Mar 10 at 19:48
2000mroliver2000mroliver
644
$endgroup$
1
$begingroup$
Yes, because the ball is compact.
$endgroup$
– Zeekless
Mar 10 at 19:50
$begingroup$
The ball is compact, so you can use this result.
$endgroup$
– angryavian
Mar 10 at 19:50
$begingroup$
The topology-level property is that continuous images of compact sets are compact. Thus in a metric space the continuous image of a compact set is, in particular, bounded.
$endgroup$
– Ian
Mar 10 at 19:51
$begingroup$
But since the set of a ball of radius 1 around the zero vector is an open set, couldn't a function be continuous on it but unbounded?
$endgroup$
– 2000mroliver
Mar 10 at 19:54
$begingroup$
On the open ball, yes. We generally distinguish between the open ball and the closed ball explicitly because of this kind of thing. (Also, in general a closed ball in a metric space is not always compact, but it is in $mathbbR^n$ or $mathbbC^n$.)
$endgroup$
– Ian
Mar 10 at 19:57
|
show 1 more comment
$begingroup$
Is a function on a ball of radius 1 around the zero vector bounded if it is continuous?
If yes/no, why is that?
general-topology functional-analysis
asked Mar 10 at 19:48
2000mroliver2000mroliver
644
$endgroup$
Is a function on a ball of radius 1 around the zero vector bounded if it is continuous?
If yes/no, why is that?
general-topology functional-analysis
general-topology functional-analysis
asked Mar 10 at 19:48
2000mroliver2000mroliver
644
asked Mar 10 at 19:48
2000mroliver2000mroliver
644
asked Mar 10 at 19:48
2000mroliver2000mroliver
644
asked Mar 10 at 19:48
2000mroliver2000mroliver
644
asked Mar 10 at 19:48
2000mroliver2000mroliver
644
644
1
$begingroup$
Yes, because the ball is compact.
$endgroup$
– Zeekless
Mar 10 at 19:50
$begingroup$
The ball is compact, so you can use this result.
$endgroup$
– angryavian
Mar 10 at 19:50
$begingroup$
The topology-level property is that continuous images of compact sets are compact. Thus in a metric space the continuous image of a compact set is, in particular, bounded.
$endgroup$
– Ian
Mar 10 at 19:51
$begingroup$
But since the set of a ball of radius 1 around the zero vector is an open set, couldn't a function be continuous on it but unbounded?
$endgroup$
– 2000mroliver
Mar 10 at 19:54
$begingroup$
On the open ball, yes. We generally distinguish between the open ball and the closed ball explicitly because of this kind of thing. (Also, in general a closed ball in a metric space is not always compact, but it is in $mathbbR^n$ or $mathbbC^n$.)
$endgroup$
– Ian
Mar 10 at 19:57
|
show 1 more comment
1
$begingroup$
Yes, because the ball is compact.
$endgroup$
– Zeekless
Mar 10 at 19:50
$begingroup$
The ball is compact, so you can use this result.
$endgroup$
– angryavian
Mar 10 at 19:50
$begingroup$
The topology-level property is that continuous images of compact sets are compact. Thus in a metric space the continuous image of a compact set is, in particular, bounded.
$endgroup$
– Ian
Mar 10 at 19:51
$begingroup$
But since the set of a ball of radius 1 around the zero vector is an open set, couldn't a function be continuous on it but unbounded?
$endgroup$
– 2000mroliver
Mar 10 at 19:54
$begingroup$
On the open ball, yes. We generally distinguish between the open ball and the closed ball explicitly because of this kind of thing. (Also, in general a closed ball in a metric space is not always compact, but it is in $mathbbR^n$ or $mathbbC^n$.)
$endgroup$
– Ian
Mar 10 at 19:57
1
1
$begingroup$
Yes, because the ball is compact.
$endgroup$
– Zeekless
Mar 10 at 19:50
$begingroup$
Yes, because the ball is compact.
$endgroup$
– Zeekless
Mar 10 at 19:50
$begingroup$
The ball is compact, so you can use this result.
$endgroup$
– angryavian
Mar 10 at 19:50
$begingroup$
The ball is compact, so you can use this result.
$endgroup$
– angryavian
Mar 10 at 19:50
$begingroup$
The topology-level property is that continuous images of compact sets are compact. Thus in a metric space the continuous image of a compact set is, in particular, bounded.
$endgroup$
– Ian
Mar 10 at 19:51
$begingroup$
The topology-level property is that continuous images of compact sets are compact. Thus in a metric space the continuous image of a compact set is, in particular, bounded.
$endgroup$
– Ian
Mar 10 at 19:51
$begingroup$
But since the set of a ball of radius 1 around the zero vector is an open set, couldn't a function be continuous on it but unbounded?
$endgroup$
– 2000mroliver
Mar 10 at 19:54
$begingroup$
But since the set of a ball of radius 1 around the zero vector is an open set, couldn't a function be continuous on it but unbounded?
$endgroup$
– 2000mroliver
Mar 10 at 19:54
$begingroup$
On the open ball, yes. We generally distinguish between the open ball and the closed ball explicitly because of this kind of thing. (Also, in general a closed ball in a metric space is not always compact, but it is in $mathbbR^n$ or $mathbbC^n$.)
$endgroup$
– Ian
Mar 10 at 19:57
$begingroup$
On the open ball, yes. We generally distinguish between the open ball and the closed ball explicitly because of this kind of thing. (Also, in general a closed ball in a metric space is not always compact, but it is in $mathbbR^n$ or $mathbbC^n$.)
$endgroup$
– Ian
Mar 10 at 19:57
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If the ball is open, then you cannot say that the function is bounded. For example, consider the function $f : B rightarrow mathbbR$ defined on the open ball $B$ by
$$textfor all xin B, quad f(x) = frac1$$
This function is not bounded.
If however your function is defined on a whole finite-dimensional space, then it is also defined on the closed ball which is compact. Hence it is continuous on a compact, so it is bounded on it, and a fortiori, it is bounded on the open ball.
answered Mar 10 at 20:25
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
add a comment |
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$begingroup$
If the ball is open, then you cannot say that the function is bounded. For example, consider the function $f : B rightarrow mathbbR$ defined on the open ball $B$ by
$$textfor all xin B, quad f(x) = frac1$$
This function is not bounded.
If however your function is defined on a whole finite-dimensional space, then it is also defined on the closed ball which is compact. Hence it is continuous on a compact, so it is bounded on it, and a fortiori, it is bounded on the open ball.
answered Mar 10 at 20:25
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
add a comment |
$begingroup$
If the ball is open, then you cannot say that the function is bounded. For example, consider the function $f : B rightarrow mathbbR$ defined on the open ball $B$ by
$$textfor all xin B, quad f(x) = frac1$$
This function is not bounded.
If however your function is defined on a whole finite-dimensional space, then it is also defined on the closed ball which is compact. Hence it is continuous on a compact, so it is bounded on it, and a fortiori, it is bounded on the open ball.
answered Mar 10 at 20:25
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
add a comment |
$begingroup$
If the ball is open, then you cannot say that the function is bounded. For example, consider the function $f : B rightarrow mathbbR$ defined on the open ball $B$ by
$$textfor all xin B, quad f(x) = frac1$$
This function is not bounded.
If however your function is defined on a whole finite-dimensional space, then it is also defined on the closed ball which is compact. Hence it is continuous on a compact, so it is bounded on it, and a fortiori, it is bounded on the open ball.
answered Mar 10 at 20:25
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
If the ball is open, then you cannot say that the function is bounded. For example, consider the function $f : B rightarrow mathbbR$ defined on the open ball $B$ by
$$textfor all xin B, quad f(x) = frac1$$
This function is not bounded.
If however your function is defined on a whole finite-dimensional space, then it is also defined on the closed ball which is compact. Hence it is continuous on a compact, so it is bounded on it, and a fortiori, it is bounded on the open ball.
answered Mar 10 at 20:25
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 20:25
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 20:25
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 20:25
TheSilverDoeTheSilverDoe
3,837112
3,837112
add a comment |
add a comment |
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1
$begingroup$
Yes, because the ball is compact.
$endgroup$
– Zeekless
Mar 10 at 19:50
$begingroup$
The ball is compact, so you can use this result.
$endgroup$
– angryavian
Mar 10 at 19:50
$begingroup$
The topology-level property is that continuous images of compact sets are compact. Thus in a metric space the continuous image of a compact set is, in particular, bounded.
$endgroup$
– Ian
Mar 10 at 19:51
$begingroup$
But since the set of a ball of radius 1 around the zero vector is an open set, couldn't a function be continuous on it but unbounded?
$endgroup$
– 2000mroliver
Mar 10 at 19:54
$begingroup$
On the open ball, yes. We generally distinguish between the open ball and the closed ball explicitly because of this kind of thing. (Also, in general a closed ball in a metric space is not always compact, but it is in $mathbbR^n$ or $mathbbC^n$.)
$endgroup$
– Ian
Mar 10 at 19:57