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For a set to be continuous, it's contour sets must be closed. Since we can define $$ \succsim x^0 = \{x, x^0 \in X: x \succsim x^0 \} $$ and $$ \precsim x^0 = \{x, x^0 \in X: x \precsim x^0\}$$

it can be seen that $\succsim x^0$ is the upper contour set of $X$ and $\precsim x^0$ is the lower contour set of $X$. Therefore, if $\succsim$ is continuous on $X$, it must the case that both $\succsim x^0$ and $\precsim x^0$ are closed sets.

Is my answer correct?

Edit: definition of continuous preference I'm using:

Continuous preference - MGW

Pedro Cunha
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    The argument looks circular to me. What is your definition of continuous preferences? – Regio May 13 '20 at 03:31
  • I'll add it to the question, @Regio – Pedro Cunha May 13 '20 at 03:47
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    Then your answer is not correct. You claim "For a set to be continuous, it's contour sets must be closed", but provide no proof of such a statement. Note that being able to write the definition of the upper contour sets and lower contour sets of $x_0$ is not a sufficient argument to show they are closed. – Regio May 13 '20 at 04:02
  • I see. I thought I could simply use the definition provided by MGW and start from there. If it's not too much trouble, could you point me in the right direction of the proof? – Pedro Cunha May 13 '20 at 04:03
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    MGW is almost spelling out the complete proof, you want to take a Cauchy sequence inside the set $\succsim x^0$ and show that its limit point is also in the set. Remember that a set is closed if it contains all its limit points. – Regio May 13 '20 at 04:08
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    Doesn't the text in your screenshots prove it? (The sentence starting with "An equivalent way to state ...") –  May 13 '20 at 05:48
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    I guess it does, @KennyLJ. I'm sorry for asking something rather obvious, it's just that it's my first time dealing with MGW and it's been somewhat overwhelming. I appreciate the tips and candor. I'll read it until I can understand the proof. Thanks! – Pedro Cunha May 13 '20 at 11:38

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