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Consider the following (Lp). \begin{align}\min&\quad\mathbf{c} \cdot \mathbf{x}\\ \text{s.t. }&\quad A\mathbf{x} \geq \mathbf{b} \\ &\quad\mathbf{x} \geq \mathbf{0}\end{align}

If $S$ is the feasible set of (Lp), Prove that the following definitions are equivalent.

  1. (Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v → −∞ \;$ as $v → ∞.$

  2. (Lp) is unbounded if $\; ∀x ∈ S,\; ∃ \,y ∈ S\;;\; c y < cx.$

Proving 1 $\Rightarrow$ 2 is an easy result from the definition of limits.

Proving 2 $\Rightarrow$ 1 is not easy! There is no guarantee that you can find a sequence whose limit goes to $−∞$. The fact that we have a decreasing sequence of $cx$ for $x ∈ S$ does not necessarily imply (1). Maybe this sequence doesn't get less than a specific number, i.e, it has an infimum in $\Bbb R$.

Remark : If all coordinates of $c$ are positive, since we know that $ x \ge 0$ it results that for any $x ∈ S$, $cx \ge 0.$ In this case of course the limit of any such sequence is not $−∞$.

What are your thoughts? Are these 2 definitions really equivalent?

Consider the following (Lp). \begin{align}\min&\quad\mathbf{c} \cdot \mathbf{x}\\ \text{s.t. }&\quad A\mathbf{x} \geq \mathbf{b} \\ &\quad\mathbf{x} \geq \mathbf{0}\end{align}

If $S$ is the feasible set of (Lp), Prove that the following definitions are equivalent.

  1. (Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v → −∞ \;$ as $v → ∞.$

  2. (Lp) is unbounded if $\; ∀x ∈ S,\; ∃ \,y ∈ S\;;\; c y < cx.$

Proving 1 $\Rightarrow$ 2 is an easy result from the definition of limits.

Proving 2 $\Rightarrow$ 1 is not easy! There is no guarantee that you can find a sequence whose limit goes to $−∞$. The fact that we have a decreasing sequence of $cx$ for $x ∈ S$ does not necessarily imply (1). Maybe this sequence doesn't get less than a specific number, i.e, it has an infimum in $\Bbb R$.

Remark : If all coordinates of $c$ are positive, since we know that $ x \ge 0$ it results that for any $x ∈ S$, $cx \ge 0.$ In this case of course the limit of any such sequence is not $−∞$.

Are these 2 definitions really equivalent?

Consider the following (Lp). \begin{align}\min&\quad\mathbf{c} \cdot \mathbf{x}\\ \text{s.t. }&\quad A\mathbf{x} \geq \mathbf{b} \\ &\quad\mathbf{x} \geq \mathbf{0}\end{align}

If $S$ is the feasible set of (Lp), Prove that the following definitions are equivalent.

  1. (Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v → −∞ \;$ as $v → ∞.$

  2. (Lp) is unbounded if $\; ∀x ∈ S,\; ∃ \,y ∈ S\;;\; c y < cx.$

Proving 1 $\Rightarrow$ 2 is an easy result from the definition of limits.

Proving 2 $\Rightarrow$ 1 is not easy! There is no guarantee that you can find a sequence whose limit goes to $−∞$. The fact that we have a decreasing sequence of $cx$ for $x ∈ S$ does not necessarily imply (1). Maybe this sequence doesn't get less than a specific number, i.e, it has an infimum in $\Bbb R$.

Remark : If all coordinates of $c$ are positive, since we know that $ x \ge 0$ it results that for any $x ∈ S$, $cx \ge 0.$ In this case of course the limit of any such sequence is not $−∞$.

What are your thoughts? Are these 2 definitions really equivalent?

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SecretAgentMan
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Consider the following (Lp). \begin{align}\min&\quad\mathbf{c} \cdot \mathbf{x}\\ \text{s.t. }&\quad A\mathbf{x} \geq \mathbf{b} \\ &\quad\mathbf{x} \geq \mathbf{0}\end{align}

If $S$ is the feasible set of (Lp), Prove that the following definitions are equivalent.

  1. (Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v → −∞ \;$ as $v → ∞.$

  2. (Lp) is unbounded if $\; ∀x ∈ S,\; ∃ \,y ∈ S\;;\; c y < cx.$

Proving 1 $\Rightarrow$ 2 is an easy result from the definition of limits.

Proving 2 $\Rightarrow$ 1 is not easy! There is no guarantee that you can find a sequence whose limit goes to $−∞$. The fact that we have a decreasing sequence of $cx$ for $x ∈ S$ does not necessarily imply (1). Maybe this sequence doesn't get less than a specific number, i.e, it has an infimum in $\Bbb R$.

Remark : If all coordinates of $c$ are positive, since we know that $ x \ge 0$ it results that for any $x ∈ S$, $cx \ge 0.$ In this case of course the limit of any such sequence is not $−∞$.

What are your thoughts? Are these 2 definitions really equivalent?

Consider the following (Lp). \begin{align}\min&\quad\mathbf{c} \cdot \mathbf{x}\\ \text{s.t. }&\quad A\mathbf{x} \geq \mathbf{b} \\ &\quad\mathbf{x} \geq \mathbf{0}\end{align}

If $S$ is the feasible set of (Lp), Prove that the following definitions are equivalent.

  1. (Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v → −∞ \;$ as $v → ∞.$

  2. (Lp) is unbounded if $\; ∀x ∈ S,\; ∃ \,y ∈ S\;;\; c y < cx.$

Proving 1 $\Rightarrow$ 2 is an easy result from the definition of limits.

Proving 2 $\Rightarrow$ 1 is not easy! There is no guarantee that you can find a sequence whose limit goes to $−∞$. The fact that we have a decreasing sequence of $cx$ for $x ∈ S$ does not necessarily imply (1). Maybe this sequence doesn't get less than a specific number, i.e, it has an infimum in $\Bbb R$.

Remark : If all coordinates of $c$ are positive, since we know that $ x \ge 0$ it results that for any $x ∈ S$, $cx \ge 0.$ In this case of course the limit of any such sequence is not $−∞$.

What are your thoughts? Are these 2 definitions really equivalent?

Consider the following (Lp). \begin{align}\min&\quad\mathbf{c} \cdot \mathbf{x}\\ \text{s.t. }&\quad A\mathbf{x} \geq \mathbf{b} \\ &\quad\mathbf{x} \geq \mathbf{0}\end{align}

If $S$ is the feasible set of (Lp), Prove that the following definitions are equivalent.

  1. (Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v → −∞ \;$ as $v → ∞.$

  2. (Lp) is unbounded if $\; ∀x ∈ S,\; ∃ \,y ∈ S\;;\; c y < cx.$

Proving 1 $\Rightarrow$ 2 is an easy result from the definition of limits.

Proving 2 $\Rightarrow$ 1 is not easy! There is no guarantee that you can find a sequence whose limit goes to $−∞$. The fact that we have a decreasing sequence of $cx$ for $x ∈ S$ does not necessarily imply (1). Maybe this sequence doesn't get less than a specific number, i.e, it has an infimum in $\Bbb R$.

Remark : If all coordinates of $c$ are positive, since we know that $ x \ge 0$ it results that for any $x ∈ S$, $cx \ge 0.$ In this case of course the limit of any such sequence is not $−∞$.

Are these 2 definitions really equivalent?

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TheSimpliFire
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Consider the following (Lp). $$ \min \;\; \mathbf{c} \cdot \mathbf{x}\\ \text{s.t. } A\mathbf{x} \geq \mathbf{b} \\ \mathbf{x} \geq \mathbf{0} \\ $$ If\begin{align}\min&\quad\mathbf{c} \cdot \mathbf{x}\\ \text{s.t. }&\quad A\mathbf{x} \geq \mathbf{b} \\ &\quad\mathbf{x} \geq \mathbf{0}\end{align}

If $S$ is the feasablefeasible set of (Lp), Prove that the following definitions are equivalent.

  1. (Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v → −∞ \;$ as $v → ∞.$

  2. (Lp) is unbounded if $\; ∀x ∈ S,\; ∃ \,y ∈ S\;;\; c y < cx.$

Proving 1 $\Rightarrow$ 2 is an easy result from the definitondefinition of limits.

Proving 2 $\Rightarrow$ 1 is not easy! There is no guarantee that you can find a sequence whose limit goes to $−∞$. The fact that we have a decreasing sequence of $cx$ for $x ∈ S$ does not necessarily imply (1). Maybe this sequence doesn't get less than a specific number, i.e, it has an infimum in $\Bbb R$.

Remark : If all coordinates of $c$ are positive, since we know that $ x >= 0$$ x \ge 0$ it results that for any $x ∈ S$, $cx >= 0.$$cx \ge 0.$ In this case of course the limit of any such sequence is not $−∞$.

What are your thoughts? Are these 2 definitions really equivalent?

Consider the following (Lp). $$ \min \;\; \mathbf{c} \cdot \mathbf{x}\\ \text{s.t. } A\mathbf{x} \geq \mathbf{b} \\ \mathbf{x} \geq \mathbf{0} \\ $$ If $S$ is the feasable set of (Lp), Prove that the following definitions are equivalent.

  1. (Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v → −∞ \;$ as $v → ∞.$

  2. (Lp) is unbounded if $\; ∀x ∈ S,\; ∃ \,y ∈ S\;;\; c y < cx.$

Proving 1 $\Rightarrow$ 2 is an easy result from the definiton of limits.

Proving 2 $\Rightarrow$ 1 is not easy! There is no guarantee that you can find a sequence whose limit goes to $−∞$. The fact that we have a decreasing sequence of $cx$ for $x ∈ S$ does not necessarily imply (1). Maybe this sequence doesn't get less than a specific number, i.e, it has an infimum in .

Remark : If all coordinates of $c$ are positive, since we know that $ x >= 0$ it results that for any $x ∈ S$, $cx >= 0.$ In this case of course the limit of any such sequence is not $−∞$.

What are your thoughts? Are these 2 definitions really equivalent?

Consider the following (Lp). \begin{align}\min&\quad\mathbf{c} \cdot \mathbf{x}\\ \text{s.t. }&\quad A\mathbf{x} \geq \mathbf{b} \\ &\quad\mathbf{x} \geq \mathbf{0}\end{align}

If $S$ is the feasible set of (Lp), Prove that the following definitions are equivalent.

  1. (Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v → −∞ \;$ as $v → ∞.$

  2. (Lp) is unbounded if $\; ∀x ∈ S,\; ∃ \,y ∈ S\;;\; c y < cx.$

Proving 1 $\Rightarrow$ 2 is an easy result from the definition of limits.

Proving 2 $\Rightarrow$ 1 is not easy! There is no guarantee that you can find a sequence whose limit goes to $−∞$. The fact that we have a decreasing sequence of $cx$ for $x ∈ S$ does not necessarily imply (1). Maybe this sequence doesn't get less than a specific number, i.e, it has an infimum in $\Bbb R$.

Remark : If all coordinates of $c$ are positive, since we know that $ x \ge 0$ it results that for any $x ∈ S$, $cx \ge 0.$ In this case of course the limit of any such sequence is not $−∞$.

What are your thoughts? Are these 2 definitions really equivalent?

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