# Thread: closed and convex sets

1. ## closed and convex sets

How does one proof the closeness in this case?

the questions is show that $(C)_{\mathbf{v}}$ is closed and convex if $C$ is.

$(C)_{\mathbf{v}}$ = {x : x = $\mathbf{v}$ + y for some y in $C$ }

To prove that $(C)_{\mathbf{v}}$ is convex i did this:

Let $x_{1}, x_{2}$ be in $(C)_{\mathbf{v}}$, then

$x_{1} = \mathbf{v} + y_{1}$ and
$x_{2} = \mathbf{v} + y_{2}$ , $y_{1}, y_{2}$ in $C$.

thus,
$x = \mu x_{1} + (1 - \mu)x_{2}$
$x =\mathbf{v} + \mu y_{1} + (1 - \mu)y_{2}$

but we know that $C$ is convex so
$\mu y_{1} + (1 - \mu)y_{2}$ is in $C$

thus
$x =\mathbf{v} + \mu y_{1} + (1 - \mu)y_{2}$ is in $(C)_{\mathbf{v}}$.

but how does one prove that $C_{\mathbf{v}$ is closed?

2. Originally Posted by mtmath
How does one proof the closeness in this case?

the questions is show that $(C)_{\mathbf{v}}$ is closed and convex if $C$ is.

$(C)_{\mathbf{v}}$ = {x : x = $\mathbf{v}$ + y for some y in $C$ }

To prove that $(C)_{\mathbf{v}}$ is convex i did this:

Let $x_{1}, x_{2}$ be in $(C)_{\mathbf{v}}$, then

$x_{1} = \mathbf{v} + y_{1}$ and
$x_{2} = \mathbf{v} + y_{2}$ , $y_{1}, y_{2}$ in $C$.

thus,
$x = \mu x_{1} + (1 - \mu)x_{2}$
$x =\mathbf{v} + \mu y_{1} + (1 - \mu)y_{2}$

but we know that $C$ is convex so
$\mu y_{1} + (1 - \mu)y_{2}$ is in $C$

thus
$x =\mathbf{v} + \mu y_{1} + (1 - \mu)y_{2}$ is in $(C)_{\mathbf{v}}$.

but how does one prove that $C_{\mathbf{v}$ is closed?

It'd be great if you'd tell us what is C: is this a vector/metric/norm/in general, topological space or what?

Tonio

3. Originally Posted by tonio
It'd be great if you'd tell us what is C: is this a vector/metric/norm/in general, topological space or what?

Tonio
Actually the whole question reads as:

Let $(C)_{\mathbf{v}}$ denote the translation of a set $C$ by vector $\mathbf{v}$, i.e.,

$(C)_{\mathbf{v}} \triangleq \{x: x = \mathbf{v}$ + y for some y $\in C\}.$

Show that $(C)_{\mathbf{v}}$ is also closed and convex if $C$ is.

4. Originally Posted by mtmath
Actually the whole question reads as:

Let $(C)_{\mathbf{v}}$ denote the translation of a set $C$ by vector $\mathbf{v}$, i.e.,

$(C)_{\mathbf{v}} \triangleq \{x: x = \mathbf{v}$ + y for some y $\in C\}.$

Show that $(C)_{\mathbf{v}}$ is also closed and convex if $C$ is.

I understood the question from the beginning: where are C and v taken from?! What's the question's context?

Tonio

5. Originally Posted by tonio
I understood the question from the beginning: where are C and v taken from?! What's the question's context?

Tonio
I meant to say this is an exercise after a chapter on projections Onto convex sets. $C$ and $\mathbf{v}$ are not defined on the question.

6. Originally Posted by mtmath
I meant to say this is an exercise after a chapter on projections Onto convex sets. $C$ and $\mathbf{v}$ are not defined on the question.

C must be defined as some kind of topological structure, otherwise I can't see how "closedness" can

be defined...

Tonio

7. Originally Posted by tonio
C must be defined as some kind of topological structure, otherwise I can't see how "closedness" can

be defined...

Tonio
$C$ is not defined anywhere in the question. So should i take my answer as a complete answer to the question?