**YOU WANT ME DO NBHM OR CSIR QUESTIONS?. PLEASE COMMENT BELOW.**

1) 3,

2) 4,

3) 5,

d) 7.

**I have given a detailed solution below so that the same idea can be used in similar problems. Do read completely.**

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following**

**the blog by email.**

**Visit my blog every day and share the solutions with friends. Thank you.**

**Solution:**Let $(X,\tau_1)$ and $(X,\tau_2)$ be topological spaces on a set $X$. A bijection $f: X \to X$ is said to be homeomorphism from $(X,\tau_1)$ to $(X,\tau_2)$ if $f$ is a continuous open map. In particular, the image and inverse image of an open set is open.

Given $X = \{1,2,3\}$. Note that,

**in any topology on $X$ a proper open set has cardinality either $1$ to $2$.**

**option 1**: (

**Number of topologies with three open sets**) Let $(X,\tau)$ be a topology on $X = \{1,2,3\}$ with three open sets. Then $\tau = \{X,\phi, U\}$ where $U$ is a subset of cardinality either $1$ or $2$. There are three subsets of $X$ with cardinality $1$ and three subsets with cardinality $2$. Any of these $6$ subsets can play the role of $U$ in $\tau$ to form a topology with three open sets. Hence there are $6$ topologies on $X$ with $3$ open sets. If $|U| = 1$ then we call such topologies of type I and if $|U| = 2$ then we call them as type $II$.

Claim: Any two topologies of type $I$ are isomorphic. Let $\tau_1 = \{X,\phi,U_1\}$ and $\tau_2 = \{X, \phi, U_2\}$ be two such topologies $U_1 = \{a\}$ and $U_2 = \{b\}$ where $a,b \in \{1,2,3\}$. Now, any permutation of $X= \{1,2,3\}$ which maps $a$ to $b$ is a homeomorphism between $(X,\tau_1)$ to $(X,\tau_2)$.

Claim: Any two topologies of type $II$ are isomorphic. Let $\tau_1 = \{X,\phi,U_1\}$ and $\tau_2 = \{X, \phi, U_2\}$ be two such topologies $U_1 = \{a,b\}$ and $U_2 = \{c,d\}$ where $a,b,c,d \in \{1,2,3\}$. Now, any permutation of $X= \{1,2,3\}$ which maps $\{a,b\}$ to $\{c,d\}$ is a homeomorphism between $(X,\tau_1)$ to $(X,\tau_2)$.

Claim: A topology of type $I$ and a topology of type $II$ cannot be homeomorphic. This follows from the following observation.

**observation:**Since $f$ is a homeomorphism, for an openset $U$ of $X$, we have $|f(U)| = |U|$ and $|f^{-1}(U)| = |U|$.

**Therefore there are two homeomorphism classes of topologies on $X$ with three open sets and these classes are given by type $I$ and type $II$ defined above.**

**option 2:(Number of topologies with four open sets)**

We use the same ideas as above. Let $(X,\tau = \{X,\phi,U_1,U_2\})$ be a topology on $X$ with four open sets. Assume $U_1 = \{a\}$ then $U_2$ cannot be an open set with a single point because then their union will also be in $\tau$ which will give us the fifth open set. So $U_2$ has to be a set with two elements. Similarly, if $|U_1| = 2$ then $|U_2| = 1$. Without loss of generality, we assume that $|U_1| = 1$. Now there are two cases. $U_1 \subseteq U_2$ we call such topologies type $I$ and $U_1 \nsubseteq U_2$ we call such topologies type $II$. There are

**six**topologies of type $I$ and there are

**three**topologies of type $II$. So there are

**nine**topologies on $X$ with four open sets. Among these topologies, like in the previous option, one can show that any two topologies of the same type (type I or type II) are homeomorphic and two topologies of different types are not homeomorphic.

**Hence there are two homeomorphism classes of topologies with four open sets in $X$.**

**option 3:(Number of topologies with five open sets)**

Let $(X,\tau = \{X,\phi,U_1,U_2,U_3\})$ be a topology on $X$ with five open sets.

Assume $U_1 = \{a\}$ then $U_2$ can be an open set with a single point or two points.

**Assume $U_2$ is singleton**then $U_3$ has to be $U_1 \cup U_2$ as we have only five open sets in $\tau$. We call such topologies type $I$. There are three topologies of this type.

**Assume $U_2$ has two elements.**

**Claim: $a \in U_2$.**If not, $U_2 = \{b,c\}$ with $b \ne a$

and $c \ne a$. Lets check the choices for $U_3$

suppose $|U_3| =1$. If $a \notin U_3$ then $U_3 \cap U_2$ is a sixth open set in $\tau$. So $U_3 = U_1$. This shows that $|U_3| =2$. By our assumption $U_2$ doesn't contain $a$ which means that $U_3$ should contains $a$ otherwise $U_2 = U_3$. If $a \in U_3$ then $U_3 \cup U_2$ is a sixth open set in $\tau$. Contradiction. So if $a \notin U_2$ there is no choice for $U_3$. Hence we have $a \in U_2$.

**Claim: $a \in U_3$.**Same argument as above.

**This shows that $\tau = \{X,\phi,\{a\},\{a,b\},\{a,c\}\}$.**We call such topologies of type $II$ and there are three of them (a=1 or 2 or 3). Hence, including type $I$, there are a total of

**six**topologies on $X$ with five open sets. Among these topologies, like in the previous option, one can show that any two topologies of the same type are homeomorphic and two topologies of different types are not homeomorphic.

**Hence there are two homeomorphism classes of topologies with four open sets in $X$.**

**option 4:(Number of topologies with seven open sets)**

We claim that there is no topology on $X$ with $7$ open sets.

Let $\tau$ be a topology on $X = \{1,2,3\}$.

Claim:

**If there are at least seven open sets in $\tau$ then $\tau$ is the discrete topology on $X$**(which is nothing but the power set of $X$ where all the subsets are open).

**This shows that there are no topology on $X$ with $7$ open sets.**

proof: It is enough to prove that all the singleton subsets of $X$ are in $\tau$. Since $\tau$ is closed under arbitrary union, $\tau$ will contain all the subsets if it contains all the singleton subsets. We have $|\tau| = 7$ and hence either all the singleton subsets of $X$ are in $\tau$ or all the two elements subsets of $X$ are in $\tau$. In the first case we are done and if all the two-element subsets of $X$ are in $\tau$ then by taking intersection among them we can generate all the singletons and again we are done.

**here**for more problems.

**Share to your groups:**

**FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.**

Kudos for your efforts... really nice work...

ReplyDeleteThis will be easier if we think : In how many ways we can draw topologies with the gn requirements in 3 point set?

Also for the last option : discrete topology has 8 open sets... To find a topology with 7 open sets, u need to miss exactly one from discrete topology ... which will cause trouble...

Thank you folk for your encouragement. Well... about topology with seven element your claim is perfect. I hope I have explain what causes the trouble. Can you add more details about your first claim? Thanks. Do follow the blog by email. Cheers.

ReplyDeleteThank you... I have sent the ans by mail... Hearty wishes... Keep it up.

ReplyDeleteThank you very much

ReplyDeleteWelcome friend. Please follow the blog regularly and follow by email to get regular notifications. Follow by email option is given below and and also top left. All the best.

ReplyDelete