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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?

Emptymind
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    Please copy the relevant definitions over from the other question. You have to show that the empty set and the full set are open, that an arbitrary union of opens is open, and that a finite intersection of opens is again open. Where do you get stuck? Furthermore, I don't think the proof-writing tag is relevant here; the tag wiki says: "This tag should ... not be used to ask for a proof of a statement." – Aldoggen Jan 27 '20 at 13:21

2 Answers2

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.

William Elliot
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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).

Henno Brandsma
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