# Thread: Discrete metrics (basic topology)

1. ## Discrete metrics (basic topology)

Let X be a set donated by the discrete metrics
d(x,y) = 0 if x=y,
1 if x̸=y.

Show that a subset Y of X is compact iff this set is closed

2. Originally Posted by hebby
Let X be a set donated by the discrete metrics
d(x,y) = 0 if x=y,
1 if x̸=y.

Show that a subset Y of X is compact iff this set is closed
I don't understand. Maybe I am speaking out of turn here, but isn't every subset of $\displaystyle X$ closed? For, to claim that it wasn't closed would be to say that the subset doesn't contain one of it's limit points but no point in $\displaystyle X$ has a limit point (as can be seen by taking a neighborhood of radius one-half). Did you mean finite?

3. Yes finite, im sorry I was reading another question at the same time

4. Originally Posted by hebby
Yes finite, im sorry I was reading another question at the same time
A finite point set is always compact. Conversely, suppose that $\displaystyle Y$ was compact, but infinite. Then $\displaystyle Y$ must have a limit point....so

THIS IS ROUGH ON PURPOSE. Clean it up.

5. Another way to see this is to consider an open cover by open balls with radius $\displaystyle \frac 12$.

6. Originally Posted by putnam120
Another way to see this is to consider an open cover by open balls with radius $\displaystyle \frac 12$.
Another way to do what? This surely isn't compact, but you have given one example, not a proof.

7. First, notice $\displaystyle \left(x_n\right)_{n\in\mathbb{N}}\subset X$ converges iff it becomes constant after a certain $\displaystyle N\in\mathbb{N}$. With that, every subset is closed.

Now, im not sure about this excercise. I think $\displaystyle Y$ must be finite... if not we may choose $\displaystyle X=\mathbb{N}$, $\displaystyle Y=X-\{1\}$ and the sequence $\displaystyle x_n=n+1.$ that doesn't have any convergent sub-sequence 'cause it never become constant.

8. Y is finite

9. HI

What do mean by constant?...could you explain this? thanks

10. Originally Posted by Drexel28
Another way to do what? This surely isn't compact, but you have given one example, not a proof.
I was just giving a place to start from. Obviously finite sets are compact. Then using this you can show that any compact set must be finite.

11. Originally Posted by hebby
HI

What do mean by constant?...could you explain this? thanks
$\displaystyle \exists A,N:x_n=A,\forall n\geq N.$

If $\displaystyle Y$ is finite, then $\displaystyle (x_n)\subset Y$ have at least one term infinitly repeated and here u have your sub-sequence for compactness.
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