Question:
Let $A=\left [ 0,1 \right )$ and $X=\left [ 0,1 \right ]$.
Prove that if $\left ( X,\alpha \right )$ is a topological space and if A is given a subspace topology then the inclusion map i is continuous.
Attempt:
By definition of the inclusion map:
i:$\left ( A,\beta \right )\rightarrow \left ( X,\alpha \right )$.
$a \mapsto \left ( a \right )i=a$
Recall:
A function $f:\left ( A,\beta \right )\rightarrow \left ( X,\alpha \right )$ is called continuous IFF for every $a \in \beta$
we have $f^{-1}\left ( a \right ) \in \alpha$
It suffice to show that the pre-image of every element in $\beta$ is in $\alpha$.
The subspace topology $\beta$ on A is a topology so the empty set $\varnothing \in \beta$.
Hence, by definition of the inclusion map, $\left ( \varnothing \right )i^{-1}=\varnothing \in \alpha$
$A=\left [ 0,1 \right ) \in \beta$.
$\left ( A \right )i^{-1}=A \in \alpha$ since $A \subseteq X$ and $X \in \alpha$ by definition of topology $\alpha$ on X.
I would like to ask if I am on the right track?
If I am not, any hints are appreciated.
Thanks in advance.
Best Answer
You just have to keep in mind what the subspace topology is: a subset of $A$ belongs to $\beta$ if it is the intersection of $A$ with an element of $\alpha$.
But the preimage of a set under $\iota$ is exactly its intersection with $A$. So in particular the preimage of a set in $\alpha$ is the intersection of that set with $A$, which by definition of the subspace topology $\beta$ means that it belongs to $\beta$.
So preimages of sets in $\alpha$ belong to $\beta$, which means continuity.