Given $C \subset A \subset X$, why is that $C$ is closed in $X$ if $A$ is closed, $C$ is open in $X$ if $A$ is open?

Question

I want to understand a result discussed here : Subspace of a normal space

Let $(X, \mathfrak{T})$ be a topological space.

1. Given $C \subset A \subset X$, let $C$ be a closed set in $A$, then claim that $C$ is closed in $X$ if $A$ is closed.

2. Similarly, let $C$ be an open set in $A$, then $C$ is open in $X$ if $A$ is open.

I am stuck on understanding why these are true even though it looks trivial.

1. Let $C$ be a closed set in the subspace topology of $A$. Then $C$ is the intersection of all the closed sets in $A$ for which $C$ is a subset. Since $A$ is closed, $A$ is the intersection of all the closed set in $X$ for which $A$ is a subset. Then the intersection of all the closed set in $X$ plus all the closed sets in $A$ is $C$. Therefore $C$ is closed in $X$. Is this argument sound?

2. If $C$ is open in $A$, then $C$ lies in the subspace topology of $A$, i.e. $C = U \cap A$, $U \in \mathfrak{T}$. Then $C \subset U$ or $C \subset A$. How does it follow that $C$ is open in $X$? Recall $C$ is open in $X$ if $C \in \mathfrak{T}$, what guarantee do we have just because $C \subset U, U \in \mathfrak{T}$ therefore $C \in \mathfrak{T}$?

Is there some intuitive way to see that they are true?

Answer 1

In 2, the sets $U$ and $A$ are both in $\mathfrak{T}$, i.e. they are both open sets of $X$. But the intersection of two open sets is open, so $C=U \cap A$ is open.

A similar logic will work for 1, but you must use that a set is closed if and only if its complement is open, making the argument somewhat more complicated.

Answer 2

For 1:

Let $C\subseteq A$ be closed in $(A,\mathfrak{T}|_A)$ and let $A$ be closed in $(X,\mathfrak{T})$

Then $A\setminus C$ is open in $(A,\mathfrak{T}|_A)$ by the definition of being closed.

Then, $(A\setminus C)\cap A$ is open in $(X,\mathfrak{T})$ by definition of the induced topology.

Furthermore, $X\setminus A$ is open in $(X,\mathfrak{T})$ by the fact that $A$ is closed.

Then $X\setminus C = (X\setminus A)\cup (A\setminus C)=(X\setminus A)\cup ((A\setminus C)\cap A)$ is the union of two open sets and is therefore open.

Therefore, $C$ is closed by definition.