In the past paper of topology, it says that the set given by:
$$X^* = \{\{a, d\}, \{b\}, \{c\}, \{e\}\}$$
with the endowed topology;
$$\mathcal{T} := \{X, ∅, \{a, b\}, \{c\}, \{a, b, c\}, \{d, e\}, \{a, b, d, e\}, \{c, d, e\}\}$$ $$X := \{a, b, c, d, e\}$$
Why is $X^*$ not connected and not Hausdorff?
From my understanding of Hausdorff, there are some elements $x_1,$ and $x_2$ such that there exists open sets $U$ and $V$ where $U\cap V = \emptyset$, $x_1\in U$ and $x_2\in V$. And from $X^*$, we can find two open sets from the topology generated by $X^* \rightarrow( \mathcal{T}_{X^*}$) , but even from the topology generated from $X^*$ I cant see why it is not Hausdorff and not connected?
From the topology generated by $X^*$ the two open sets can be $\{c\}$ and $\{\{a,d\},\{b\}, \{e\}\}$, but this is Hausdorff since it is disjoint open sets, so I am confused. So why is it NOT Hausdorff?
Can anyone help me see why it is not connected and Hausdorff?
So you have a set $X^* = \{\{a,d\},\{b\},\{c\},\{e\}\}$, of subsets of the space $X = \{a,b,c,d,e\}$. It so happens that $X^*$ doesn't form a topology by itself, because $X$ is not closed under union, for example. So we find the smallest topology containing $X^*$, by taking all such possible unions and finite intersections. This topology we denote by $\tau_{X^*}$.
It turns out, the $\mathcal T$ that you have written in the question, is not correct. It turns out $\mathcal T = \tau_{Y}$ for a different subset $Y = \{\{a,b\},\{c\},\{d,e\}\}$!
Indeed, the topology that will be generated by $X^*$, call it $\tau_{X^*}$ will be made as follows :
first, take $X^*$ itself, and take all possible unions and intersections of pairs of elements from $X^*$. This gives rise to the subsets : $\{a,b,d\},\{a,c,d\},\{a,d,e\},\{b,c\},\{b,e\},\{c,e\}$, and any two different subsets of $X^*$ are disjoint, so add in $\emptyset$.
Now, repeat by taking all possible unions of elements of the new sets which have been formed in the previous step. Proceed till you are sure nothing new is formed.
This procedure is good in the case where the number of elements of $X^*$ is small, which is the case here.
Once you do this, you will land up with the following topology.
$$ \tau_{X^*} = X^* \cup \{ \emptyset,X,\{a,b,d\},\{a,c,d\},\{a,d,e\},\{b,c\},\{b,e\},\{c,e\},\{a,c,d,e\},\\\{a,b,c,e\},\{a,b,c,d\},\{a,b,d,e\},\{b,c,e\}\} $$
which you can verify satisfies all the conditions (it is difficult to write this set down immediately as a beginner : it requires some practice to get quicker).
With the confusion of generating topology behind us, we now proceed to see why $X^*$ is not connected and not Hausdorff.
Actually, the answers are sort of clear. Why is that ? It is because, we have literally written down all the open sets above, and so verifying properties depending on these open sets is now made easier.
For example, to show that $X$ is disconnected with the topology $T_{X^*}$, one simply needs to find two disjoint open sets whose union is $X$. A brief look at the list above tells us that $\{a,c,d\}$ and $\{b,e\}$ are disjoint open sets whose union is $X$. Hence, $X$ is disconnected.
To see that $X$ is not Hausdorff, we take the two points $a$ and $d$. If $X$ was Hausdorff, then we could find two disjoint open sets, one containing $a$ and the other containing $d$. Again, a look at the list will tell you that every open set either contains both $a$ and $d$ , or contains neither of them. Therefore, you can't find a pair of open sets seperating $a$ and $d$, since the open set containing $a$ must also contain $d$, for example.
Hence, $X$ is not Hausdorff, since you can't find a pair of open sets separating $a$ and $d$.
That's all right. But surely, the person who wrote this answer did not have to go through writing the entire topology $\tau_{X^*}$ and then scratching through umpteen odd sets to find a counterexample. Also, he/she surely did not figure out by magic that $a$ and $d$ are the two elements which cannot be separated by open sets : it would take him too much time if he tried to brute force.
So here's the not-so-hidden secret : There is no need to write down the entire list of open sets to figure out whether the space is connected or not. It's actually enough to look at $X^*$ itself!
Let's look at $X^*$. It contains four disjoint sets $\{a,b\},\{c\},\{d\},\{e\}$, whose union is $X$. Therefore, the topology generated by $X^*$ contains $X^*$, and hence these sets also. But they are disjoint and their union is $X$, so $X$ is not connected!
Similarly, all of the elements of $X^*$ either contain both $a$ and $d$, or don't contain both! The same will also happen with the generating topology. Hence, the candidates $a,d$ were proposed and verified for non-Hausdorffness.
The following definitions and results I copy from Wikipedia. I will add comments.
For example, $X^*$ is a basis for $\tau_{X^*}$, by definition of generated topology.
Here are the key results governing the base , and making the emphasized statement in the yellow box above more prominent and precise.
If $B$ generates $T_B$, then $T_B$ is Hausdorff if and only if for each pair of unequal points $x\neq y$, we can find two disjoint sets $U,V \in B$ (not $T_B$!) such that $x \in U,y\in V$.
$T_B$ is disconnected if and only if $B$ contains a subcover of $X$ by disjoint sets.
For example, you can't find two open sets in $X^*$ which separate $a$ and $d$. Similarly, $X^*$ itself is a cover of $X$ by open sets. Hence the results give the required conclusions.
You may also verify that the accidentally constructed $\tau_Y$ is also not Hausdorff and not connected.
On the other hand, something like $Z = \{\{a,b\},\{c\}\}$, and taking the topology generated by this $\tau_Z$, it is not Hausdorff, but is connected, since (briefly) any open set or it's complement contains $d$, so either the open set or its complement is the entire set.
That's should hopefully complete the discussion of this problem. But I would like to add a few words regarding your comment on a previous question, which directed me to this question.
Look at connectedness : it says that there are open sets $U,V$ such that $U \cap V = \emptyset$ and $U \cup V = X$. Put another way, there is an open set $W \neq X,W \neq \emptyset$ such that $W^c$ is also open. Therefore, connectedness can be fulfilled by the inclusion of precisely one set.
On the other hand, Hausdorffness says : for each pair of points, you want open sets separating these points. This requires many open sets, but there is no requirement of each of them other than to be disjoint to a certain number of other sets.
In short, connectedness can be fulfilled by one set, which Hausdorffness is fulfilled by many open sets acting in tandem.
Seeing the requirements, in general, one does not see where the two correlate at all. However, in special situations such as the finite complement topology, there may possibly be connections between the two of them.
For example, in the finite complement topology, if $X$ is disconnected, it is finite, but then under f.c.t every subset of $X$ is open so of course $X$ is Hausdorff, so here, non-connected implies Hausdorff. Similarly, if $X$ is connected then it is non-Hausdorff. So here, $X$ is disconnected if and only if it is Hausdorff, so there is a clear connection between the two properties.
However, in general, there's no reason why the two should correlate.