# Subspace Topology

Let $X$ denote the set $[-1,1] \cup \{0'\}$. The following collection $S$ of subsets of $X$ is a basis for a topology $F$ on $X$: $S$ consists of subsets $U$ of $X$ such that 1) $U$ is an open subset of $[-1,1]$ in the usual topology or 2) there is an open subset $V$ of $[-1,1]$ which contains $0$ and $U=V-\{0\}+\{0'\}$. This is easy to show.

Show that the subspace topology on $[-1,1]$ induced from the topology $F$ on $X$ is the usual topology. Do I just show that an open interval is an open ball?

Let $g:X\to Y$ be a continuous map, where $Y$ is a metric space. Show that $g(0)= g(0')$. I'm not quite sure how to show this one.

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What do you mean by $0'$? – gnometorule Feb 19 '13 at 18:09
@gnometorule: A second copy of $0$: this is a subspace of the line with two origins. – Brian M. Scott Feb 19 '13 at 18:19
@gnometorule: Works fine in Firefox, Chrome, and Opera for me. The target is within the page <en.wikipedia.org/wiki/Non-Hausdorff_manifold>;. – Brian M. Scott Feb 19 '13 at 18:26

Since each open subset of $[-1,1]$ is again open in $X$, then the subspace topology will be finer than the standard topology. To show that the subspace topology is coarser than (so the same as) the standard topology, it suffices to show that $[-1,1]\cap U$ is open in $[-1,1]$ in the standard topology whenever $U$ is a basis element for the topology on $X$. Two quick cases should do the trick.

Fact: Suppose $Y$ is a metric space with metric $d$, and that $Z$ is any topological space with no isolated points (like $X$). Then $f:Z\to Y$ is continuous at a point $z\in Z$ if and only if for each $\epsilon>0,$ there is a neighborhood $U$ of $z$ in $Z$ such that $d(w,z)<\epsilon$ whenever $w\in U-\{z\}$. (It's a good exercise to prove that.)

We call $U-\{z\}$ (with $U,z$ as in the fact above) a "punctured neighborhood of $z$ in $Z$". Observe that every punctured neighborhood of $0$ in $X$ is a punctured neighborhood of $0'$ in $X$, and vice versa. Then use the above fact with a triangle-inequality style argument to see that $d\bigl(g(0),g(0')\bigr)<3\epsilon$ for any $\epsilon>0,$ so $d\bigl(g(0),g(0')\bigr)=0,$ and so $g(0)=g(0').$

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Hints

Part 1, show that you don't get any additional open sets in [-1,1] by removing $0$ from a normal open set.

Part 2, any open set containing $0'$ contains elements arbitrarily close to $0$.

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