Question: > 
> Question:
> 
> 
> Let $X$ be a set. Let $\tau$ consists of $\varnothing$ , and any set $U \subseteq X$ such that $X\setminus U$ is either finite or countable.
> Show that $\tau$ is a topology on $X$.
> 
> 
> 

Looking only for $\mathbf{hints}$.

Let $\tau=\left \{ \varnothing,U \right \}$.
Obviously, $X \subseteq X$ so $X \in \tau$

Now, $\tau=\left \{ \varnothing, \left \{ U\_{i} \right \}\_{i \in I} \right \}$ with $X \in \left \{ U\_{i} \right \}\_{i \in I}: \exists i \in I$

By hypothesis, $X\setminus U$ is countable so it is expected that $\exists$ a bijection

$f:\mathbb{Z}^{+}\rightarrow X\setminus U$

I would like to take this further. My intuition says that if I can show that the elements in $U$ are open sets, then the elements in $U$ are elements of $\tau$ which then satisfies one of the axiom for topological space.
However, I am unable to do so.

Any Hint is appreciated.

Thanks in advance.

Answer: 
As you wrote [in your comment](https://math.stackexchange.com/questions/1895793/simple-interest-problems#comment3891102_1895793) you have
$x+y+z=2379$. 

You also have $1.1x=1.15y=1.2z$. From this you can express all variables using one of them. For example, you have
\begin{align\*}
y&=\frac{1.1}{1.15}x=\frac{22}{23}x\\
z&=\frac{1.1}{1.2}x=\frac{11}{12}x
\end{align\*}

Plugging this into the equation $x+y+z=2379$ you get
\begin{align\*}
x+\frac{22}{23}x+\frac{11}{12}x&=2379\\
\frac{276+264+253}{276}x&=2379\\
\frac{793}{276}x&=2379\\
x&=\frac{276\cdot2379}{793}\\
x&=276\cdot3\\
x&=828
\end{align\*}

Now you can check whether the conditions of the original problem are fulfilled.
We know that:

* $x+y+z=2379$
* $1.05^2x=1.05^3y=1.05^4z$

---

We can easily replace $x$ and $y$ in the equation $x+y+z=2379$, because:

* $1.05^2x=1.05^4z \implies x=1.05^2z$
* $1.05^3y=1.05^4z \implies y=1.05^1z$

---

Therefore:

* $x+y+z=2379 \implies 1.05^2z+1.05^1z+z=2379 \implies z=\frac{2379}{3.1525}$

Therefore:

* $y=1.05^1z \implies y=1.05^1\cdot\frac{2379}{3.1525}=\frac{2497.95}{3.1525}$
* $x=1.05^2z \implies x=1.05^2\cdot\frac{2379}{3.1525}=\frac{2622.8475}{3.1525}$

---

Summary:

* $x=\frac{2622.8475}{3.1525}\approx832$
* $y=\frac{2497.95}{3.1525}\approx792$
* $z=\frac{2379}{3.1525}\approx755$