Totally disconnected implies T1: Difference between revisions

Latest revision as of 20:20, 13 January 2012

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., totally disconnected space) must also satisfy the second topological space property (i.e., T1 space)
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Statement

Any totally disconnected space is a T1 space.

Facts used

Proof

Given: A totally disconnected space $X$ , a point $x\in X$ .

To prove: ${\overline {\{x\}}}=\{x\}$ .

Proof: We prove this by noting that the closure ${\overline {\{x\}}}$ is irreducible by Fact (1), hence connected by Fact (2). Hence, because $X$ is totally disconnected, it must be the singleton subset $\{x\}$ .

Revision as of 20:19, 13 January 2012 (view source) Vipul (talk \| contribs) (→‎Statement) ← Older edit		Latest revision as of 20:20, 13 January 2012 (view source) Vipul (talk \| contribs) (→‎Proof)
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	'''To prove''': <math>\overline{\{ x \}} = \{ x \}</math>.		'''To prove''': <math>\overline{\{ x \}} = \{ x \}</math>.

	'''Proof''': We prove this by noting that the closure <math>\overline\{ x \}</math> is connected~~, on account of being an [[irreducible space]]~~. ~~Thus~~, ~~by the definition of~~ totally disconnected, it must be a singleton subset.		'''Proof''': We prove this by noting that the closure <math>\overline{\{ x \}}</math> is irreducible by Fact (1), hence connected by Fact (2). Hence, because <math>X</math> is totally disconnected, it must be the singleton subset <math>\{ x \}</math>.