# Show that if $\succsim$ is continous on $X$, then the sets $\precsim (x^0)$ and $\succsim (x^0)$ are closed

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.

• 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. May 13, 2020 at 4:02
• 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. May 13, 2020 at 4:08