1.791. Let𝑥∈int𝑆.Thus𝑆is a neighborhood of𝑥.Therefore,𝑇⊇𝑆is aneighborhood of𝑥, so that𝑥is an interior point of𝑇.2. Clearly, if𝑥∈𝑆, then𝑥∈𝑇⊆𝑇. Therefore, assume𝑥∈𝑆∖𝑆which impliesthat𝑥is a boundary point of𝑆. Every neighborhood of𝑥contains other pointsof𝑆⊆𝑇. Hence𝑥∈𝑇.1.80Assume that𝑆is open. Every𝑥∈𝑆has a neighborhood which is disjoint from𝑆𝑐. Hence no𝑥∈𝑆is a closure point of𝑆𝑐.𝑆𝑐contains all its closure points and istherefore closed.Conversely, assume that𝑆is closed.Let𝑥be a point its complement𝑆𝑐.Since𝑆is closed and𝑥 /∈𝑆,𝑥is not a boundary point of𝑆.This implies that𝑥has aneighborhood𝑁which is disjoint from𝑆, that is𝑁⊆𝑆𝑐. Hence,𝑥is an interior pointof𝑆𝑐. This implies that𝑆𝑐contains only interior points, and hence is open.
This is the end of the preview.
access the rest of the document.