Closure is linear subspace

Let X be normed linear space, Y a subspace of X. The closure of Y, \bar{Y}, is a linear subspace of X.

Proof:
We use the “sequential” equivalent definition of closure, rather than the one using open balls: \bar Y is the set of all limits of all convergent sequences of points in Y. Let z_1,z_2\in \bar{Y}, \alpha\in\mathbb{R}. There is a sequence (a_n) in Y such that a_n\to z_1. Similarly there is a sequence (b_n) in Y which converges to z_2.

Then (a_n+b_n) is a sequence in Y that converges to z_1+z_2. (\alpha a_n) is a sequence in Y that converges to \alpha z_1.

Advertisements

About mathtuition88

http://mathtuition88.com
This entry was posted in math and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s