(DISCONTINUED) #2: Munkres Problem Set 2.13.4

2.13.4

Problem:

  1. If \{\mathcal{T}_\alpha\} is a family of topologies on X, show that \bigcap\mathcal{T}_\alpha is a topology on X. Is \bigcup\mathcal{T}_\alpha a topology on X?
  2. Let \{\mathcal{T}_\alpha\} be a family of topologies on X. Show that there is a unique smallest topology on X containing all collections \mathcal{T}_\alpha, and a unique largest topology contained in all \mathcal{T}_\alpha.
  3. If X=\{a,b,c\}, let \mathcal{T}_1=\{\emptyset,X,\{a\},\{a,b\}\} and \mathcal{T}_2=\{\emptyset,X,\{a\},\{b,c\}\}. Find the smallest topology containing \mathcal{T}_1 and \mathcal{T}_2, and the largest topology contained in \mathcal{T}_1 and \mathcal{T}_2.

Answer:

  1. \bigcap\mathcal{T}_\alpha is a topology on X due to the fact that each \mathcal{T}_\alpha is a topology on X and if \{\mathcal{T}_\alpha\} is empty, then \bigcap\mathcal{T}_\alpha is the discrete topology on X. For \bigcup\mathcal{T}_\alpha, let X=\{a,b,c\} and define the topologies \mathcal{T}_1=\{\emptyset,X,\{a\},\{a,b\}\} and \mathcal{T}_2=\{\emptyset,X,\{a\},\{b,c\}\}. If we take an intersection of \mathcal{T}_1 and \mathcal{T}_2, then we are presented with a situation where \{a,b\}\cap\{b,c\}=\{b\}\notin\mathcal{T}_1\cup\mathcal{T}_2. Therefore showing that \bigcup\mathcal{T}_\alpha is not a topology on X. \blacksquare
  2. We must note that there exists a topology on X containing all the collections \mathcal{T}_\alpha (discrete topology). Thus, there exists a topology on X contained in all X. Therefore denoted as \mathcal{T}_l=\bigcap\mathcal{T}_\alpha to be the largest topology on X. For the smallest topology on X that contains all the collections \mathcal{T}_\alpha, we take the intersection of all topologies that satisfy the property of containing the union of all the \mathcal{T}_\alpha such that \mathcal{T}_s=\bigcap_{\bigcup\mathcal{T}_\alpha\subset F}F where there exists at least one F. Therefore I claim that \mathcal{T}_l is the largest topology and \mathcal{T}_s is the smallest topology on X. To show the uniqueness of \mathcal{T}_s, let F be a topology that contains all the \mathcal{T}_\alpha. \mathcal{T_s} is the intersection of X and therefore \mathcal{T}_s\subset F. Then suppose \mathcal{T}'_s fits the criteria of \mathcal{T}_s, then \mathcal{T}'_s\subseteq\mathcal{T}_s and \mathcal{T}_s\subseteq\mathcal{T}'_s. And therefore \mathcal{T}_s=\mathcal{T}'_s, thus \mathcal{T}_s is unique. The uniqueness proof of \mathcal{T}_l is similar. \blacksquare
  3. The smallest topology containing \mathcal{T}_1 and \mathcal{T}_2 is the union of \mathcal{T}_1 and \mathcal{T}_2 such that \mathcal{T}_1\cup\mathcal{T}_2=\{\emptyset,X,\{a\},\{a,b\},\{b,c\},\{b\}\} for which \{b\} has to be thrown in where \{b\}=\{a,b\}\cap\{b,c\}. The largest topology contained in \mathcal{T}_1 and \mathcal{T}_2 is the intersection such that \mathcal{T}_1\cap\mathcal{T}_2=\{\emptyset,X,a\}. \blacksquare
Advertisements

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