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This a lengthy post with several parts; I would appreciate any help offered on any of the parts.

Let $X$ be a set equipped with any topology $\tau_1$, let $\tau_2$ be the cocountable topology (i.e. a subset $U$ of $X$ is open if either $U=\emptyset$ or $X\setminus U$ is finite or countable), and let $\tau=\tau_1\lor \tau_2$ be defined as the subsets $W\subseteq X$ such that for every $x \in W$, there exist subsets $U,V \subseteq X$ with $x \in U$, $x \in V$, $U \in \tau_1$, $V \in \tau_2$, and $U \cap V \subseteq W$.

1.) If $X$ is finite or countable, show that $\tau$ is the discrete topology.

To show that $\tau$ is discrete, I need to show that any subset $W \subseteq X$ is open with respect to $\tau$. It suffices to show that every singleton set in $X$ is open, since by the definition of a topology this implies that any union of singleton sets, and thus any subset, will be open. Consider $\{x\} \subseteq X$. I'm not sure how to set this up to get started.

2.) Prove that a point $p \in X$ is a limit point of a set $E \subseteq X$ with respect to $\tau$ if and only if $p$ is a condensation point of $E$ with respect to $\tau_1$.

For the forward direction, suppose that $p$ is a limit point of $E$ with respect to $\tau$. Then, by definition, for every neighborhood $N$ of $p$, there exists some $n \in N$ such that $n \neq p$. I don't know where to go from here.

For the other direction, suppose that $p$ is a condensation point of $E$ with respect to $\tau_1$. Then, by definition, for every subset $U \subseteq X$ with $U \in \tau_1$ and $p \in U$, $E \cap U$ is uncountable. From this definition, it is clear that $p$ is a limit point of $E$ with respect to $\tau_1$. I am not sure how to get from here to the fact that $p$ is a limit point of $E$ with respect to $\tau$.

3.) When is a set $E \subseteq X$ dense in $X$ with respect to $\tau$?

I'm not sure what the appropriate statement to prove here should be. I know when a set $E \subseteq X$ is dense with respect to $\tau_2$. Specifically, if $X$ is countable, then $X$ is the only dense set with respect to $\tau_2$. If $X$ is uncountable, then $E \subseteq X$ is dense with respect to $\tau_2$ iff $E$ is uncountable. How does this relate to which sets are dense with respect to $\tau$?

EDIT: If $X$ is uncountable, a set $E \subseteq X$ is dense with respect to $\tau$ if and only if every point $p' \in X \setminus E$ is a condensation point of $E$ with respect to $\tau_1$.

My attempt at a proof: For the forward direction, suppose $E \subseteq X$ is dense in $X$ with respect to $\tau$. Then, by definition, for every nonempty open set $U \subseteq X$, $E \cap U \neq \emptyset$. That is, there exists some $x \in E \cap U$ for every open set $U \subseteq X$. We want to conclude that every point $p' \in X \setminus E$ is a limit point of $E$ with respect to $\tau$, but I'm not sure how to fill in the details here.

For the reverse direction, let $p' \in X \setminus E$. Suppose that $p'$ is a condensation point of $E$ with respect to $\tau_1$. Then, by part 2, $p'$ is a limit point of $E$ with respect to $\tau$. How does this get us to the conclusion that $E$ is dense?

4.) Suppose that $(X,\tau_1)$ is Hausdorff. Is $(X, \tau)$ Hausdorff? Prove or give a counterexample.

By definition, $(X,\tau_1)$ is Hausdorff implies that for every $x,y \in X$, with $x \neq y$, there exist $U,V \in \tau_1$ such that $x \in U$, $y \in V$, and $U \cap V = \emptyset$. My gut tells me the answer is no, but I cannot think of a counterexample, so I am not sure.

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  • $\begingroup$ When you say, $U\subseteq \tau_1, V\subseteq \tau_2$ should that be $U\in \tau_1, V\in \tau_2$? $\endgroup$ – saulspatz Mar 30 '18 at 3:05
  • $\begingroup$ Yes! I will edit $\endgroup$ – mathqueen459 Mar 30 '18 at 3:06
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Claim. $\tau_1\subseteq\tau$ and $\tau_2\subseteq\tau$.

For $W\in\tau_1$, choose $U=W\in\tau_1$ and $V=X\in\tau_2$ (as saulspatz showed in the answer above) for every $x\in W$. Then $x\in U\cap V=W\cap X=W$, and so $W\in\tau$ by the definition of $\tau=\tau_1\vee\tau_2$. In a similar way, we can show $\tau_2\subseteq\tau$. (Choose $U=X\in\tau_1$ and $V=W\in\tau_2$ for $W\in\tau_2$.)

1) If $X$ is finite or countable, show that $\tau$ is the discrete topology.

As saulspatz mentioned, $\tau_2$ is nothing but the discrete topology. Since $\tau$ contains all elements of $\tau_2$ by the claim above, $\tau$ is also the discrete topology.

2) Prove that a point $p \in X$ is a limit point of a set $E \subseteq X$ with respect to $\tau$ if and only if $p$ is a condensation point of $E$ with respect to $\tau_1$.

We may assume that $X$ is uncountable. Otherwise (2) follows immediately from (1).

($\Rightarrow$) [Use contraposition.] Suppose $p$ is not a condensation point of $E$ w.r.t. $\tau_1$. Then there is an open set $U\in\tau_1$ containing $p$ such that $U\cap E$ is not uncountable, (i.e., finite or countable). Set $V\equiv X\setminus\bigl[U\cap (E-\{p\})\bigr]\in\tau_2$. Notice that $p\in U\in\tau_1$ and $p\in V\in\tau_2$. Since $\tau$ contains both $\tau_1$ and $\tau_2$ by the claim above, we have $p\in U\cap V\in\tau$, but $$ (U\cap V)\cap(E-\{p\}) = \bigl\{ X\setminus\bigl[U\cap (E-\{p\})\bigr] \bigr\} \cap \bigl[U\cap(E-\{p\})\bigr]=\varnothing $$ Thus $p$ is not a limit point of $E$ w.r.t. $\tau$.

($\Leftarrow$) Suppose $p$ is a condensation point of $E$ w.r.t. $\tau_1$. Let $W$ be an open set in $\tau$ containing $p$. Then, by the definition of $\tau$, there are two open sets $p\in U\in\tau_1$ and $p\in V\in\tau_2$ such that $U\cap V\subseteq W$. Since $p$ is a condensation point of $E$ w.r.t. $\tau_1$, $U\cap E$ is uncountable, and so is $U\cap(E-\{p\})$. If $V\cap(U\cap(E-\{p\})=\varnothing$, then $V$ does not contain uncountably many points of $E-\{p\}$. It contradicts to $V\in\tau_2$. Therefore $p$ is a limit point of $E$ w.r.t. $\tau$, since $W$ contains a point in $E$ other than $p$. $$ W\cap(E-\{p\}) \supseteq (U\cap V)\cap(E-\{p\}) \neq \varnothing $$

3) When is a set $E \subseteq X$ dense in $X$ with respect to $\tau$?

We also assume that $X$ is uncountable. Otherwise $X$ itself is the only dense set. $$ \begin{align*} \text{$E\subseteq X$ is dense in $\tau$} &\iff \text{Every $p\notin E$ is a limit point of $E$ w.r.t. $\tau$} \\ &\iff \text{Every $p\notin E$ is a condensation point of $E$ w.r.t. $\tau_1$} \\ &\iff \text{Is there any simple condition? (I have no idea yet.)} \end{align*} $$

Note. OP said that "If $X$ is uncountable, then $E\subseteq X$ is dense with respect to $\tau_1$ iff $E$ is uncountable." But it is not true. The set of rational numbers $E=\mathbb{Q}$ is a countable dense subset of the set of real numbers $X=\mathbb{R}$.

4) Suppose that $(X,\tau_1)$ is Hausdorff. Is $(X, \tau)$ Hausdorff? Prove or give a counterexample.

Of course $X$ is Hausdorff w.r.t. $\tau$, because $\tau$ is finer than $\tau_1$. (It is always true that if a topology $\tau$ on $X$ is Hausdorff then so is the finer topology $\tau'\supseteq\tau$ on $X$.)

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  • $\begingroup$ My bad on part 3! That should be with respect to $\tau_2$, not $\tau_1$. I will edit the question @ChoF $\endgroup$ – mathqueen459 Apr 4 '18 at 4:43
  • $\begingroup$ I'm struggling with the proof of part 3. Will you look at my edit that includes my proof attempt? $\endgroup$ – mathqueen459 Apr 6 '18 at 5:34
  • $\begingroup$ @mathqueen459 You tried to show that $E\subseteq X$ is dense in $\tau$ if and only if every point $p\in X\setminus E$ is a condensation point of $E$ w.r.t. $\tau_1$. See my answer of (3). The second $\iff$ comes from (2). The first $\iff$ comes from the definition of "dense". Do you want to show the first $\iff$? $\endgroup$ – ChoF Apr 7 '18 at 2:36
  • $\begingroup$ Yes, I'm wanting to show the first implication, which should come from the definition of dense somehow. $\endgroup$ – mathqueen459 Apr 7 '18 at 21:40
  • $\begingroup$ @mathqueen459 It's simple. $E$ is dense if and only if the closure of $E$ is $E$ itself. Since the closure is the union of $E$ and the set of limit points of $E$, it suffices to show that all points in $X-E$ is a limit points of $E$. $\endgroup$ – ChoF Apr 8 '18 at 23:44
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For the first part, if $X$ is countable or finite, then $\tau_2$ is the discrete topology -- the complement of every set is finite or countable.
So then given any $W\subseteq X,$ we can take $U=X,$ $V=W$.

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