Fix a set $X$ with at least 2 elements and $p\in X$. Let $\mathcal{T}_p = \{ U \subset X \mid p \notin U \} \cup \{ X \}$.
Show that $\mathcal{T}_p$ is a topology and denote this topological space by $X$.
Is $X$ connected?
Is $X$ Hausdorff?
Let $(x_n)$ be the sequence with $x_n = a$ for all $n$ for some $a \in X$. For what values of $a$ does $(x_n)$ converge? If $(x_n)$ converges, where does it converge to?
My work.
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$X, \emptyset \in \mathcal{T}_p$.
Suppose $U_1 \subset X$ and $U_2 \subset X$ and $p \notin U_1$ and $p \notin U_2$ then $p \notin U_1 \cup U_2 \subset X$ so $U_1 \cup U_2 \in \mathcal{T}_p$ Hence unions of elements of $\mathcal{T}_p$ are in $\mathcal{T}_p$.
Suppose $U_1 \subset X$ and $U_2 \subset X$ and $p \notin U_1$ and $p \notin U_2$ then $p \notin U_1 \cap U_2 \subset X$ so $U_1 \cap U_2 \in \mathcal{T}_p$ Hence intersections of elements of $\mathcal{T}_p$ are in $\mathcal{T}_p$.
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$X$ is connected since the only set in $\mathcal{T}_p$ containing $p$ is the entire set $X$, and hence there cannot exist a separation.
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$X$ is not Hausdorff since it is connected
I think (1) and (2) are right, I think (3) is wrong and I don't know what to do for (4).
Best Answer
This is almost correct, but for the union part write it up for arbitrary unions, not just finite ones. Make explicit why $\emptyset \in \mathcal{T}_p$.
The reason is essentially correct, make the write-up more formal: suppose $X = A \cup B$ is a separation of $X$. One of the sets contains $p$ etc...
This has a bogus argument: the reals are connected and very Hausdorff. Think of the same reason why $X$ is connected (neighbourhoods of $p$) and use that $X$ has another element besides $p$...
Just use the definition of convergence. And consider for limits all $x \neq p$ and $p$ separately. What is a typical open neighbourhood of $x \neq p$ and $p$? Does the sequence stay inside the neighbourhood or not?