(Conceptual) Continuity of binary relation $\succsim$ and definition using contour sets

Question

Some background information: $\succsim$ is a binary relation that represents preference between two goods. $\succsim$ means "x is at least as good as y." Continuity of this relation is defined to be that if it is preserved in limits: if for any pair of sequences $(x^n,y^n)$ converging to x and y respectively and $x^n\succsim y^n$ $\forall n$, then $x\succsim y$. Also, $x\in\mathbb{R^L_+}$. So, X is a set of possible goods we can choose and these goods are L-vectors.

I am trying to understand that an equivalent way of stating "a preference relation that is continuous on $X$" is to say "the upper and lower contours sets are closed". What I don't understand is authors in a book says the following:

The continuity of $\succsim$ on X implies that for any sequence of points $\{y^n\}$ with $x\succsim y^n$ $\forall n$ and $y^n$ converging to $y$, we have $x\succsim y$. And, this implies the lower contour $L(x)=\{y\in X:x\succsim y\}$ set is closed.

My instinct was to show the lower contour set is closed is equivalent to showing the set contains all its limit points. And how do you do this? Well, in this case, it is to show whether, say the lower contour set, contains its boundary points. These points in collection is just the indifference set $I(x)=\{y\in X:y\sim x\}$.

Then, I want to show $I(x)\subset L(x)$.

Proof : (1) Suppose $x\in I(x)$.

(2) By definition, $x\succsim y$ and $y\succsim x$ $\iff x\sim y$. Hence, $I(x)=\{y\in X:y\succsim x\}\cap\{y\in X:x\succsim y\}$.

(3) Hence, $x\in L(x)$ and $I(x)\subset L(x)$. Q.E.D.

Is my approach incomplete and how should I make sense of MWG's claim above? Please help. Thank you!!

Answer 1

Continuity of this relation is defined to be that if it is preserved in limits: if for any pair of sequences $(x^n,y^n)$ converging to x and y respectively and $x^n\succsim y^n$ $\forall n$, then $x\succsim y$.

This statement easily implies

for any sequence of points $\{y^n\}$ with $x\succsim y^n$ $\forall n$ and $y^n$ converging to $y$, we have $x\succsim y$.

if you fix the constant sequence $x^n=x$ for all $n$. Then note that this statement is equivalent to saying "if $y^n\in L(x)$ and $y_n\to y$, then $y\in L(x)$." You'll see that this is exactly the definition of $L(x)$ being sequentially closed (see Wikipedia). Thus you know that the statement

the lower contour $L(x)=\{y\in X:x\succsim y\}$ set is closed.

is definitely true as long as "sequentially closed" implies "closed." But counterintuitively, this is not always the case! Topologies in which this is true are called sequential topologies. You can guarantee you are in one of these if $\mathbb L$ is at most countable since then $\mathbb R^\mathbb L$ will be first countable (see the same Wikipedia page).