Showing that the topology on a disjoint union is indeed atopology.

Question

I have the topologies given in 2 answers here Putting a topology on the disjoint union and what are the open sets. but actually I am unable to show that either one of them is a topology. could anyone help me in doing so please?

Answer 1

Let X and Y be two spaces and X + Y their
disjoint union as defined in the reference.
Show the empty set is an open set of X + Y.
Show X + Y is an open set of X + Y.

Let U and V be two open sets of X + Y.
Thus exists open within X, $U_x, V_x$
and exists open within Y, $U_y, V_y$ with
$U = (U_x×\{1\}) \cup (U_y×\{2\}), V = (V_x×\{1\}) \cup (V_y×\{2\}).$

What is the intersection of U and V?
Is it open within X + Y?

In a similar way, show any union of
open sets of X + Y is an open set of X + Y.

Answer 2

Defining $X \coprod Y$ as $\{(x,0): x \in X\} \cup \{(y,1): y \in Y\}$ and the standard injections $j_X: X \to X \coprod Y$ by $j_X(x)=(x,0)$ and $j_Y: Y \to X \coprod Y$ by $j_Y(y)=(y,1)$, the topology is defined as the final topology induced by $j_X,j_Y$ namely

$$\mathcal{T}= \{O \subseteq X \coprod Y: j^{-1}_X[O] \in \mathcal{T}_X \land j^{-1}_Y[O] \in \mathcal{T}_Y\}$$

which is easily checked to be a topology (as $\mathcal{T}_X,\mathcal{T}_Y$ are, and inverse images preserve unions and intersections).