1. ## Compactness

For a set M ⊂ R^2, we define its shadow (projection) on the x-axis to be the set in R defined as

M_x={ x: there is some y so that (x,y) ∈ M }

a) Find a set M, so that M_x is open (in R) but M is not open (in R^2).

b) Suppose K is a compact set in R^2. Prove that K_x is also compact in R.

Hint: Either prove …first "if A is open in R, then A × R is open in R^2", then start with an open cover C of K_x and create a cover for K.
Or: start with a sequence (x_n) in K_x and use that K is sequentially compact to show that (x_n) has a cluster point in K_x.

c) Find a closed set N in R^2 such that N_x is not closed (in R) .

If anyone helps me, I will be thankful.

2. a) $M=\{(x,y)|0
you can prove why

b) Taken an open cover of $K_x$ call it A. Look at B where B consists of the inverse of the projection map of each open set in A. For instance if $a \in A$ then the corresponding open set in B is $a \times \mathbb{R}$. This is an open set since it is a product of two open sets in $\mathbb{R}^2$, so we notice B is an open cover of K, a compact set, so it has a finite subcover. Then look at the projections of each of these sets and it is a finite subcover of $K_x$ and thus its compact.

I dunno how to do the fancy script A, so I apologize for the awkward notation of a representing a set and A being a collection of sets...

c) $\Delta = \{(x,x)\in \mathbb{R}^2|x\in \mathbb{R}\}$ This is the diagonal and it is probably a theorem you have done that is pretty easy to show from definitions that hausdorff iff diagonal is closed so $\Delta$ is closed in $\mathbb{R}^2$ but its projection is $\mathbb{R}$ which is open.

If you dont believe me that $\Delta$ is closed, just look at $\mathbb{R}^2-\Delta$ and see why it's open.

Hope that helps.

3. Originally Posted by Gamma
c) $\Delta = \{(x,x)\in \mathbb{R}^2|x\in \mathbb{R}\}$ This is the diagonal and it is probably a theorem you have done that is pretty easy to show from definitions that hausdorff iff diagonal is closed so $\Delta$ is closed in $\mathbb{R}^2$ but its projection is $\mathbb{R}$ which is open.

If you dont believe me that $\Delta$ is closed, just look at $\mathbb{R}^2-\Delta$ and see why it's open.
$\Delta$ is certainly closed, but its projection is $\mathbb{R}$, which is closed as well as open.

To find a closed set in $\mathbb{R}^2$ whose projection is not closed, you need to look at something like $\{(x,1/x):x>0\}$.