# What's wrong with this 'open cover' of the Koch Snowflake?

This question is to help me find peace.

First, the question of the Snowflake's compactness has been tackled here on this site:

But I'm inclined to believe it is not. The only reason Henning Makholm's argument does not fully convince me is because I believe I can come up with an open cover without finite subcover. What you do is surround each triangle with an open triangle small enough so that it doesn't intersect the 'sprouts' two levels down which sprouted out of the first iteration, and then repeat that ad infinitum. I hope this picture helps (the grey lines represents the open sets):

So, my question is why does this open cover have a finite subcover?

-
What exactly is your cover? Are you sure that you've covered the point $(1/3,0)$? – Mark McClure Feb 13 '14 at 2:44
I've only posted one side of the original triangle. The grey lines implicitly connect by a straight line on the 'inside.' And I implicitly did this pattern to each triangle. – Bryan Feb 13 '14 at 2:48

It's not an open cover. Take a point of a "sprout" that is not in open set number $n$. These can be chosen to converge (and in any case some subsequence must converge). The limit is in the snowflake but not in any of your open sets.
The curve is the image of a continuous function over the set $[0,1]$ as you can see in the answer that you post. Therefore, as compactness is preserved under continuous functions, the Koch curve is compact.