Half-space and open set

[Anonymous No.12272833]

How do I prove this?

[Anonymous No.12272845]

x -> p*x is a continuous map
R^k -> R
The preimages of open sets under continuous maps are open.

[Anonymous No.12272880]

Dot products increase when the two vectors are more similar, decrease (into the negatives) when they are more dissimilar (ie: when the arrow in k space is pointed the opposite direction.).

So this is the set of all vectors x that are less similar to vector p than some measure, B.

In a space visualization, this would draw a decision boundary (set by B) as some hyperplane cutting through k dimensional space. If vector x lies on one side of the boundary, it's in the set A; on the other side, it's not on the set A.

The number of such possible vectors is infinite, this it's an open set.

[Anonymous No.12272890]

>>12272880
** The number of vectors on either side of the decision plane is infinite.

[Anonymous No.12272928]

Anyone can use the methond used in the link to prove it pls ? https://math.stackexchange.com/questions/698638/closedness-of-the-closed-half-space

[Anonymous No.12273883]

>>12272833
Open sets are just ones that don't contain their boundaries. Show the boundary is p.x=b which by definition it doesn't contain.

[Anonymous No.12273976]

>>12272833
draw A for k=2 and k=3 gor intuition, then use the definitions, whats hard about this?

[Anonymous No.12273981]

>>12273883
>Open sets are just ones that don't contain their boundaries
Wrong

[Anonymous No.12274023]

>>12272833
you should think about the ball centered at x with radius (beta - p•x)/|p|, or something similar.

[Anonymous No.12274034]

Use Cauchy Schwartz

[Anonymous No.12274035]

>>12274023
yes, this is it
say y is in the ball. then p•y - p•x <= |p•(x-y)| <= |p||x-y| < beta - p•x. then p•y < beta.

[Anonymous No.12274076]

>>12273883
every set that contains its boundary is not open but not every non open set contains its boundary