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Continuity in Topological Spaces ... M & M, Lemma 3.2 (ii) ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
I am reading Aisling McCluskey and Brian McMaster: Undergraduate Topology, Oxford University Press, 2014... ... and am currently focused on Chapter 3: Continuity and Convergence ...

I need help in order to fully understand M & M's proof of Lemma 3.2 (ii) ...


Lemma 3.2 (plus the definition of continuity) reads as follows:



M & M - Lemma 3.2 ... .png


The proof of Lemma 3.2 (ii) reads as follows:


M & M - Proof of Lemma 3.2 (ii) ... .png


In the above proof we read the following:

" ... ... But \(\displaystyle A \subseteq f^{ -1} ( f(A) ) \subseteq f^{ -1} ( \overline{ f(A) } )\) ... ...

... ... therefore \(\displaystyle \overline{A} \subseteq f^{ -1} ( \overline{ f(A) } )\) ... ... "


Can someone please demonstrate formally and rigorously that \(\displaystyle A \subseteq f^{ -1} ( f(A) ) \subseteq f^{ -1} ( \overline{ f(A) } )\) implies that \(\displaystyle \overline{A} \subseteq f^{ -1} ( \overline{ f(A) } )\) ... ...


Help will be much appreciated ...

Peter
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,653
Leeds, UK
The proof of Lemma 3.2 (ii) reads as follows:

.
.
.

In the above proof we read the following:

" ... ... But \(\displaystyle A \subseteq f^{ -1} ( f(A) ) \subseteq f^{ -1} ( \overline{ f(A) } )\) ... ...

... ... therefore \(\displaystyle \overline{A} \subseteq f^{ -1} ( \overline{ f(A) } )\) ... ... "


Can someone please demonstrate formally and rigorously that \(\displaystyle \overline{A} \subseteq f^{ -1} ( \overline{ f(A) } )\) ... ...
This is actually a proof of (ii) $\Rightarrow$ (iii) in Lemma 3.2. So we are assuming that (ii) holds. In particular, since $\overline{ f(A) }$ is closed in $(Y,\tau')$, it follows from (ii) that $f^{ -1} ( \overline{ f(A) } )$ is closed in $(X,\tau)$. But $A$ is contained in the closed set $f^{ -1} ( \overline{ f(A) } )$, and therefore $\overline{A}$ (which is the smallest closed set containing $A$) is contained in $f^{ -1} ( \overline{ f(A) } )$.