Show the discrete topology is the only one larger than $\tau_l$

Question

$(X,\le)$ is a partially ordered set, we define $U_l(x)=\{y\ |\ y\le x\}$, and $\tau_l$ is the topology generated by $\{U_l(x)\}$. We want to prove that the discrete topology is the only on that's larger than $\tau_l$.

I don't understand this question, because I think we can take: $X=\{1,2,3\}$ with the usual ordering. We then find $\tau_l=\{\emptyset,\{1\},\{1,2\},\{1,2,3\}\}$. But we can add $\{2\}$ as an open set to obtain the topology $\{\emptyset,\{1\},\{2\},\{1,2\},\{1,2,3\}\}$, which is larger than $\tau_l$ but smaller than the discrete topology.

So am I misunderstanding something here, or is this question not correct?

@StefanHamcke Yes I think so too. It's one of the first questions from the (I believe highly respected) book 'Topology' by James Dugunji though, so I would be very surprised if it is indeed false. — user2520938, Jul 06 '15 at 16:51
@JahanClaes Ah I think I see. It says that discrete is the only one that's larger the $\tau_l$ and $\tau_r$ simultaneously. My example is no counter example since it's not large than $\tau_r$ — user2520938, Jul 06 '15 at 18:00

score 3 · Accepted Answer · answered Jul 06 '15 at 18:00

I think you've misread the question.

He asks you to prove that the discrete topology is the only topology greater than BOTH $\tau_l$ and $\tau_r$, where $\tau_r$ is the topology generated by all $U_r(x)=\{y|x\leq y\}$. For a topology to be greater than both of these topologies, it has to contain both all the right-sided intervals AND all the left-sided intervals.

In the example you've given above, any topology greater than both $\tau_l$ and $\tau_r$ would have to contain $\{1\},\{1,2\},\{1,2,3\}$ and $\{1,2,3\}, \{2,3\},\{3\}$. If you play with unions and intersections, you'll see this is the discrete topology.

Yea of course, I should have thought about this some more... Thanks. — user2520938, Jul 06 '15 at 18:01

Show the discrete topology is the only one larger than $\tau_l$

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