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I need to prove that if $\tau_1$ and $\tau_2$ are topologies of a set $X$, then $\tau_1 \cup \tau_2$ is not necessarily a topology on $X$.

I'm looking for counterexamples. I have one: consider $X=\{a,b,c\}$, and $\tau_1=\{\emptyset, X, \{a,c\}\}$ and $\tau_2=\{\emptyset, X, \{a,b\}\}$ then $\tau=\{\emptyset, X, \{a,c\}, \{a,b\}\}$, but note that: $\{a,b\} \cap \{a,c\} =\{a\}\notin \tau$, therefore $\tau$ is not a topology for $X$.

Is this correct? Some other more interesting counterexample?

Thanks for your help.

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Your example is correct! Can you please say what is your measure of "interesting"? Regards, – user21436 Apr 15 '12 at 18:32
@Kannappan Sampath, I think my example is correct. What do you think? I just want to know an example not so "trivial" (if exist). Thank you. – Hiperion Apr 15 '12 at 18:39
@JyrkiLahtonen I am sorry. I realise that it is a sufficient condition but not a necessary condition. – user21436 Apr 15 '12 at 20:57
@KannappanSampath, that's ok :-) Though the other direction is not too interesting given that in those two cases the union of the two topologies is the finer of them! – Jyrki Lahtonen Apr 15 '12 at 20:59
Yes, your example is correct, as I said. There are plenty of examples. Try to prove the following: Let (X,τ1) and (X,τ2) be topological spaces. If $\tau_1 \subseteq \tau_2$ or $\tau_2 \subseteq \tau_1$, then, $\tau_1 \cup \tau_2$ is a topology on $X$. (Thus, this is a sufficient condition but by no means, a sufficient condition as Jyrki in his previous comment indicates; In essence, I had a wrong statement, so I deleted that comment. I am sorry.) – user21436 Apr 15 '12 at 21:00
up vote 6 down vote accepted

Here is an example you might find "more interesting" (though I always like minimalist examples, like yours, very much) :

Take two topological spaces $X$ and $Y$, and on the product, the two following topologies : $$ \tau_X=\{U\times Y | U \mbox{ is an open of } X\} \\ \tau_Y=\{X\times V | V \mbox{ is an open of } Y\} $$ These both clearly are topologies, and their union is not (because $(U\times Y)\cap(X\times V)\notin\tau_X\cup\tau_Y$.

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