Let $U_{r}(a)\subset X$. For $x,y\in U_{r}(a)$ there exist a continuous function $f:[0,1]\to X$ such that its image is entirely in $U_{r}(a)$.

Question

Claim: Let $(X,d)$ be a metric space. Let $U_{r}(a)\subset X$ with $\epsilon >0$ and $a\in X$. For every tow points $x,y\in U_{r}(a)$ there exist a continuous function $f:[0,1]\rightarrow X$ such that $f(0)=x$, $ f(1)=y$ and its image lies entirely in $U_{r}(a)$.

Proof:

Define $f:[0,1]\rightarrow U_{r}(a), t\mapsto x+t(y-x)$. Then $f(0)=x$ and $f(1)=y$.

I don't get the next step:

The entire image of $f$ lies in $U_r(a)$, because from $d(x,y)<1$ and $d(y,a)<1$ it follows: $$d(f(t),a))=||f(t)-a||=||x+t(y-x)-a||=|| (1-t)(x-a)+t(y-a) ||\\ \leq || (1-t)||x-a||+t||y-a||<(1-t)+t=1 \quad\quad q.e.d.$$

I understand the equivalence transformations, but it was only shown that $d(f(t),a)<1$ holds.But the proof is not elaborated further. Actually one has to show that for every $t\in[0,1]$, $d(f(t),a)<r$ holds, right?

score 1 · Accepted Answer · answered Jan 17 '23 at 19:44

1

There is a mistake in the setup, "because from $d(x,y)<1$ and $d(y,a)<1$", which is repeated in the last inequality, $||x-a||$ and $||y-a||$ are less than $r$, not less than $1$. Thus the final inequality is $$ \dots < (1-t)r + tr = r $$

answered Jan 17 '23 at 19:44

BaronVT

13,613

"... because from $d(x,a)<r$ and $d(y,a)<r$ it follows..." would be correct? – distortedPicture Jan 17 '23 at 20:13
yep, that would be one way to correctly rewrite this – BaronVT Jan 18 '23 at 19:45

Let $U_{r}(a)\subset X$. For $x,y\in U_{r}(a)$ there exist a continuous function $f:[0,1]\to X$ such that its image is entirely in $U_{r}(a)$.

I don't get the next step:

1 Answers1