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

  • 3
    $\begingroup$ The argument looks circular to me. What is your definition of continuous preferences? $\endgroup$
    – Regio
    May 13, 2020 at 3:31
  • 3
    $\begingroup$ 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. $\endgroup$
    – Regio
    May 13, 2020 at 4:02
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    $\begingroup$ 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. $\endgroup$
    – Regio
    May 13, 2020 at 4:08
  • 2
    $\begingroup$ Doesn't the text in your screenshots prove it? (The sentence starting with "An equivalent way to state ...") $\endgroup$
    – user18
    May 13, 2020 at 5:48
  • 1
    $\begingroup$ 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! $\endgroup$ May 13, 2020 at 11:38


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